Noncommutative Ring Theory Notes

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The Cartan-Brauer-Hua theorem

Posted: April 12, 2024 in Division Rings, Noncommutative Ring Theory Notes
Tags: Cartan-Brauer-Hua Theorem, Hua's identity, multiplicative group of a division ring, normal subgroup

For a division ring $D,$ we denote by $Z(D)$ and $D^{\times}$ the center and the multiplicative group of $D,$ respectively.

Let $D_1$ be a division ring, and suppose that $D_2$ is a proper subdivision ring of $D_1,$ i.e. $D_2$ is a subring of $D_1, \ D_2 \ne D_1,$ and $D_2$ itself is a division ring. Then $D_2^{\times}$ is clearly a proper subgroup of $D_1^{\times}.$ Now, one may ask: when exactly is $D_2^{\times}$ a normal subgroup of $D_1^{\times} ?$ The Cartan-Brauer-Hua theorem gives the answer: $D_2^{\times}$ is a normal subgroup of $D_1^{\times}$ if and only if $D_2 \subseteq Z(D_1).$ In particular, $D_2$ must be a field. One side of the theorem is trivial: if $D_2 \subseteq Z(D_1),$ then $D_2^{\times}$ is obviously normal in $D_1^{\times}.$ The other side of the theorem is not trivial, but it’s not hard to prove either. The proof is a quick result of the following simple yet strangely significant identity!

Hua’s Identity (Loo Keng Hua, 1949). Let $D$ be a division ring, and let $a,b \in D$ such that $ab \ne ba.$ Then

$a=(b^{-1}-(a-1)^{-1}b^{-1}(a-1))(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))^{-1}.$

Proof. First see that since $ab \ne ba,$ the four elements $a,a-1,b,$ and $a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1)$ are all nonzero hence invertible. Now,

$a(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))=b^{-1}a-a(a-1)^{-1}b^{-1}(a-1)$

$=b^{-1}a-(a-1+1)(a-1)^{-1}b^{-1}(a-1)=b^{-1}a-b^{-1}(a-1)-(a-1)^{-1}b^{-1}(a-1)$

$=b^{-1}-(a-1)^{-1}b^{-1}(a-1). \ \Box$

Cartan-Brauer-Hua Theorem. Let $D_1$ be a division ring, and let $D_2$ be a proper subdivision ring of $D_1.$ If $D_2^{\times}$ is normal in $D_1^{\times},$ then $D_2 \subseteq Z(D_1).$

Proof. Suppose, to the contrary, that $D_2 \nsubseteq Z(D_1).$ Let $a \in D_1, b \in D_2$ such that $ab \ne ba.$ Then, since $D_2^{\times}$ is a normal subgroup of $D_1^{\times},$ all the elements

$b^{-1}, \ (a-1)^{-1}b^{-1}(a-1), \ a^{-1}b^{-1}a, \ (a-1)^{-1}b^{-1}(a-1)$

are in $D_2^{\times}$ and hence, by Hua’s identity, $a \in D_2.$ So every element of $D_1 \setminus D_2$ commutes with $b.$ Now let $c \in D_1 \setminus D_2.$ Then $ac \in D_1 \setminus D_2,$ because $a \in D_2,$ and therefore both $c,ac$ commute with $b.$ But then $a=acc^{-1}$ will also commute with $b$ and that’s a contradiction. $\ \Box$

Note. There are other proofs of the Theorem; for example, here is a simple but not very well-known one.

Composition of derivations in prime rings; a theorem of Posner

Posted: March 16, 2024 in Derivation, Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: Derivation, prime ring, semiprime ring

We remarked here that the composition of derivations of a ring need not be a derivation. In this post, we prove this simple yet interesting result that if $R$ is a prime ring of characteristic $\ne 2$ and if $\delta_1,\delta_2$ are nonzero derivations of $R,$ then $\delta_1\delta_2$ can never be a derivation of $R.$ But before getting into the proof of that, let me remind the reader of a little fact that tends to bug many students.

$n$ –Torsion Free vs Characteristic $\ne n.$ Let $R$ be a ring, and $n \ge 2$ an integer. Recall that we say that $R$ is $n$ –torsion free if $a \in R, na=0$ implies $a=0.$ We say that $\text{char}(R)=n$ if $n$ is the smallest positive integer such that $na=0$ for all $a \in R.$ It is clear that if $R$ is $n$ -torsion free, then $\text{char}(R) \ne n.$ The simple point I’d like to make here is that the converse is not always true. That will make a lot more sense if you look at its contrapositive, in fact the contrapositive of a stronger statement: $na=0$ for some $0 \ne a \in R$ does not always imply that $nR=(0).$ However, the converse is true if $R$ is prime. First, an example to show that the converse in not always true.

Example 1. Consider the ring $R=\mathbb{Z}_n \oplus \mathbb{Z},$ and $a=(1,0) \in R.$ Then $a \ne 0$ and $na=0.$ So $R$ is not $n$ -torsion free. but $\text{char}(R)=0 \ne n.$

Now let’s show that the converse is true if $R$ is prime.

Example 2. Let $n \ge 2$ be an integer and $R$ a prime ring. Suppose that $na=0$ for some $0 \ne a \in R.$ Then $nR=(0),$ i.e. $nr=0$ for all $r \in R.$

Proof. We have $(0)=(na)Rr=aR(nr)$ and so, since $a \ne 0$ and $R$ is prime, $nr=0. \ \Box$

Let’s now get to the subject of this post.

Lemma 1. Let $R$ be a ring, and let $\delta_1,\delta_2$ be derivations of $R.$ Then $\delta_1\delta_2$ is a derivation of $R$ if and only if

$\delta_1(a)b\delta_2(c)+\delta_2(a)b\delta_1(c)=0,$

for all $a,b,c \in R.$

Proof. Since $\delta_1\delta_2$ is clearly additive, it is a derivation if and only if it satisfies the product rule, i.e.

$\delta_1\delta_2(bc)=\delta_1\delta_2(b)c+b\delta_1\delta_2(c). \ \ \ \ \ \ \ (1)$

On the other hand, since $\delta_1,\delta_2$ are derivations of $R,$ we also have

$\delta_1\delta_2(bc)=\delta_1(\delta_2(b)c+b\delta_2(c))=\delta_1(\delta_2(b)c)+\delta_1(b\delta_2(c))$

$=\delta_1\delta_2(b)c+\delta_2(b)\delta_1(c)+\delta_1(b)\delta_2(c)+b\delta_1\delta_2(c). \ \ \ \ \ \ \ (2)$

So we get from $(1),(2)$ that

$\delta_1(b)\delta_2(c)+\delta_2(b)\delta_1(c)=0. \ \ \ \ \ \ \ \ (3)$

Replacing $b$ by $ab$ in $(3)$ gives

$0=\delta_1(ab)\delta_2(c)+\delta_2(ab)\delta_1(c)=(\delta_1(a)b+a\delta_1(b))\delta_2(c)+(\delta_2(a)b+a\delta_2(b))\delta_1(c)$

$=\delta_1(a)b\delta_2(c)+\delta_2(a)b\delta_1(c)+a(\delta_1(b)\delta_2(c)+\delta_2(b)\delta_1(c))$

$=\delta_1(a)b\delta_2(c)+\delta_2(a)b\delta_1(c), \ \ \ \ \ \ \ \text{by} \ (3). \ \Box$

Corollary. Let $R$ be a $2$ -torsion free semiprime ring, and let $\delta$ be a derivation of $R.$ Then $\delta^2$ is a derivation of $R$ if and only if $\delta=0.$

Proof. Suppose that $\delta^2$ is a derivation of $R$ and let $a \in R.$ Then choosing $\delta_1=\delta_2=\delta$ and $c=a$ in Lemma 1 gives $2\delta(a)b\delta(a)=0,$ for all $b \in R.$ So, since $R$ is $2$ -torsion free, $\delta(a)b\delta(a)=0,$ for all $b \in R.$ Thus $\delta(a)R\delta(a)=(0)$ and hence $\delta(a)=0,$ because $R$ is semiprime. $\ \Box$

Lemma 2. Let $R$ be a prime ring, and let $\delta$ be a derivation of $R$ such that $\delta(a)b=0$ for all $a \in R$ and some $b \in R.$ Then either $b=0$ or $\delta=0.$

Proof. Since $\delta(a)b=0$ for all $a \in R,$ we have $\delta(ca)b=0$ for all $a,c \in R$ and so

$0=\delta(ca)b=(\delta(c)a+c\delta(a))b=\delta(c)ab+c\delta(a)b=\delta(c)ab.$

So $\delta(c)Rb=(0)$ for all $c \in R$ and hence, since $R$ is prime, either $b=0$ or $\delta(c)=0$ for all $c \in R. \ \Box$

Remark. Lemma 2 remains true if we replace the condition $\delta(a)b=0$ by $b\delta(a)=0.$ The proof is similar, just this time replace $a$ by $ac.$

Theorem (Edward C. Posner, 1957). Let $R$ be a prime ring of characteristic $\ne 2,$ and let $\delta_1, \delta_2$ be derivations of $R.$ Then $\delta_1\delta_2$ is a derivation of $R$ if and only if $\delta_1=0$ or $\delta_2=0.$

Proof. First note that, by Example 2, the condition $\text{char}(R) \ne 2$ is the same as saying that $R$ is $2$ -torsion free. Now, suppose that $\delta_1\delta_2$ is a derivation of $R$ and let $a,b,c \in R.$ Applying Lemma 1 to $a, \delta_2(c)b, c$ gives

$\delta_1(a)\delta_2(c)b\delta_2(c)+\delta_2(a)\delta_2(c)b\delta_1(c)=0.$

But by the identity $(3)$ in Lemma 1, $\delta_1(a)\delta_2(c)=-\delta_2(a)\delta_1(c)$ and so the above becomes

$\delta_2(a)(\delta_2(c)b\delta_1(c)-\delta_1(c)b\delta_2(c))=0.$

Thus, by Lemma 2, either $\delta_2=0$ or $\delta_2(c)b\delta_1(c)-\delta_1(c)b\delta_2(c)=0.$ If $\delta_2=0,$ we are done. So suppose that

$\delta_2(c)b\delta_1(c)-\delta_1(c)b\delta_2(c)=0.$

Adding the above identity to the identity $\delta_1(c)b\delta_2(c)+\delta_2(c)b\delta_1(c)=0,$ which holds by Lemma 1, gives $2\delta_2(c)b\delta_1(c)=0.$ Hence $\delta_2(c)b\delta_1(c)=0$ and so $\delta_2(c)R\delta_1(c)=(0).$ Thus, since $R$ is prime, either $\delta_1=0$ or $\delta_2=0. \ \Box$

Example 3. The condition $\text{char}(R) \ne 2$ cannot be removed from the Theorem. Consider the polynomial ring $R:=\mathbb{Z}_2[x],$ and the derivation $\delta:=\frac{d}{dx}.$ Then $\delta \ne 0$ but $\delta^2=0$ is a derivation.

Note. Examples and the Corollary in this post are mine. I have also slightly simplified Posner’s proof of the Theorem.

ℚ-algebras with locally nilpotent derivations

Posted: March 15, 2024 in Derivation, Noncommutative Ring Theory Notes
Tags: Derivation, Leibniz formula, locally nilpotent

All rings in this post are commutative with identity. For the basics on derivations of rings see this post and this post.

Let $\delta$ be a derivation of a ring $R.$ Since $\delta$ is additive, $\delta(na)=n\delta(a)$ for all integers $n$ and all $a \in R.$ So every derivation of a ring is a $\mathbb{Z}$ -derivation. If $R$ is $\mathbb{Q}$ -algebra and $r=\frac{m}{n} \in \mathbb{Q},$ where $m,n$ are integers and $n \ne 0,$ then $n\delta(ra)=\delta(nra)=\delta(ma)=m\delta(a)$ and so $\delta(ra)=r\delta(a).$ Thus every derivation of a $\mathbb{Q}$ -algebra is a $\mathbb{Q}$ -derivation.

Definition. We say that a derivation $\delta$ of a ring $R$ is locally nilpotent if for every $a \in R,$ there exists a positive integer $n$ such that $\delta^n(a)=0.$

Example. Let $R$ be the polynomial ring $\mathbb{C}[x].$ Then the derivation $\delta:=\frac{d}{dx}$ is locally nilpotent because if $p(x) \in R$ has degree $n,$ then $\delta^{n+1}(p(x))=0.$

The following Theorem characterizes all $\mathbb{Q}$ -algebras $R$ for which there exists a locally nilpotent derivation $\delta$ and $a \in R$ such that $\delta(a)=1.$ The polynomial ring in the above Example gives one of those algebras since, in that example, $\delta(x)=1.$ It turns out that any such algebra is a polynomial ring!

Theorem. Let $\delta$ be a locally nilpotent derivation of a $\mathbb{Q}$ -algebra $R,$ and let $K:=\ker \delta.$ If there exists $x \in R$ such that $\delta(x)=1,$ then $x$ is transcendental over $K$ and $R=K[x].$

Proof. Suppose, to the contrary, that $x$ is algebraic over $K,$ and let $n$ be the smallest positive integer such that $\alpha_nx^n+\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1x+\alpha_0=0$ for some $\alpha_i \in K, \ \alpha_n \ne 0.$ Then, since by the product rule, $\delta(x^i)=ix^{i-1}\delta(x)=ix^{i-1},$ we have

$0=\delta( \alpha_nx^n+\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1x+\alpha_0)= \alpha_n\delta(x^n)+\alpha_{n-1}\delta(x^{n-1})+ \cdots + \alpha_1\delta(x)$

$=n\alpha_n x^{n-1}+(n-1)\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1,$

which contradicts the minimality of $n.$ So $x$ is transcendental over $K.$ We now show that $R=K[x].$ For $a \in R,$ let $\nu(a)$ be the smallest positive integer $m$ such that $\delta^m(a)=0.$ The proof is by induction over $\nu(a), \ a \in R.$ If $\nu(a)=1,$ then $\delta(a)=0$ and so $a \in K \subset K[x].$ Suppose now that $a \in R$ and $m:=\nu(a) \ge 2.$ Let

$\displaystyle y:=\sum_{n=0}^{m-1}\frac{(-1)^n\delta^n(a)x^n}{n!}=a-bx,$

where

$\displaystyle b=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}\delta^n(a)x^{n-1}}{n!}.$

So $a=y+bx$ and so we are done if we prove that $b,y \in K[x].$

Claim 1: $y \in K.$

Proof. We have

$\displaystyle \delta(y)=\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}\delta(\delta^n(a)x^n)=\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}(\delta^{n+1}(a)x^n+\delta^n(a)\delta(x^n))$

$\displaystyle =\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}(\delta^{n+1}(a)x^n+n\delta^n(a)x^{n-1})=\sum_{n=0}^{m-1}\frac{(-1)^n\delta^{n+1}(a)x^n}{n!}+\sum_{n=1}^{m-1}\frac{(-1)^n\delta^n(a)x^{n-1}}{(n-1)!}$

$\displaystyle =\sum_{n=0}^{m-2}\frac{(-1)^n\delta^{n+1}(a)x^n}{n!}+\sum_{n=0}^{m-2}\frac{(-1)^{n+1}\delta^{n+1}(a)x^n}{n!}=0$

and so $y \in K.$

Claim 2: $b \in K[x].$

Proof. By the Leibniz formula,

$\displaystyle \delta^{m-1}(b)=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}}{n!}\delta^{m-1}(\delta^n(a)x^{n-1})=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}}{n!}\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{n+k}(a)\delta^{m-k-1}(x^{n-1}).$

Now notice that $\delta^{n+k}(a)\delta^{m-k-1}(x^{n-1})=0$ for all $k,n$ because if $n+k \ge m,$ then $\delta^{n+k}(a)=0,$ and if $n+k < m,$ then $\delta^{m-k-1}(x^{n-1})=0.$ Thus $\delta^{m-1}(b)=0$ and so $\nu(b) < \nu(a)=m.$ Hence, by our induction hypothesis, $b \in K[x]. \ \Box$

Exercise. Let $\delta$ be a locally nilpotent derivation of a $\mathbb{Q}$ -algebra $R,$ and let $c \in R.$ Let $K:=\ker \delta,$ and define the map $f: R \to R$ by

$\displaystyle f(a):=\sum_{n=0}^{\infty}\frac{\delta^n(a)c^n}{n!},$

for all $a \in R.$ Show that $f$ is a $K$ -algebra homomorphism.

Note. The above Theorem is Theorem 2.8 in here. The proof I’ve given is essentially the same as the proof given in there; I just made the proof easier to follow.

Derivations on rings; the basics (2)

Posted: March 15, 2024 in Derivation, Noncommutative Ring Theory Notes
Tags: Derivation, Leibniz formula, matrix ring

See the first part of this post here. All rings in this post are assumed to have identity.

In the first part, we gave two examples of derivations on rings. We now give a couple of ways to make new derivations using given ones.

Example 1. Let $R$ be a ring, and let $c_1,c_2 \in Z(R),$ the center of $R.$ If $\delta_1,\delta_2 \in \text{Der}(R),$ then

i) $c_1\delta_1+c_2\delta_2 \in \text{Der}(R),$

ii) $\delta_1\delta_2-\delta_2\delta_1 \in \text{Der}(R).$

Proof. The first part is quite straightforward and so I’m going to leave it as an easy exercise. For the second part, we need to show that $\delta_1\delta_2-\delta_2\delta_1$ is additive and satisfies the product rule. So let $a,b \in R.$ Then

$\begin{aligned} (\delta_1\delta_2-\delta_2\delta_1)(a+b)=\delta_1\delta_2(a+b)-\delta_2\delta_1(a+b)= \delta_1(\delta_2(a)+\delta_2(b))-\delta_2(\delta_1(a)+\delta_1(b))\end{aligned}$

$=\delta_1\delta_2(a)+\delta_1\delta_2(b)-\delta_2\delta_1(a)-\delta_2\delta_1(b)=(\delta_1\delta_2-\delta_2\delta_1)(a)+(\delta_1\delta_2-\delta_2\delta_1)(b),$

which proves that $\delta_1\delta_2-\delta_2\delta_1$ is additive. Now, for the product rule, we have

$\begin{aligned}(\delta_1\delta_2-\delta_2\delta_1)(ab)=\delta_1(\delta_2(ab))-\delta_2(\delta_1(ab))=\delta_1(\delta_2(a)b+a\delta_2(b))-\delta_2(\delta_1(a)b+a\delta_1(b))\end{aligned}$

$=\delta_1(\delta_2(a)b)+\delta_1(a\delta_2(b))-\delta_2(\delta_1(a)b)-\delta_2(a\delta_1(b))$

$\begin{aligned} =\delta_1\delta_2(a)b+\delta_2(a)\delta_1(b)+\delta_1(a)\delta_2(b)+a\delta_1\delta_2(b)-\delta_2\delta_1(a)b-\delta_1(a)\delta_2(b)-\delta_2(a)\delta_1(b)-a\delta_2\delta_1(b)\end{aligned}$

$=\delta_1\delta_2(a)b+a\delta_1\delta_2(b)-\delta_2\delta_1(a)b-a\delta_2\delta_1(b)=(\delta_1\delta_2-\delta_2\delta_1)(a)b+a(\delta_1\delta_2-\delta_2\delta_1)(b),$

proving that $\delta_1\delta_2-\delta_2\delta_1$ satisfies the product rule. $\ \Box$

Remark. By Example 1, ii), the commutator of two derivations is a derivation. However, the composition of two derivations need not be a derivation. For example, let $R$ be the polynomial ring $\mathbb{C}[x]$ and consider the derivation $\delta:=\frac{d}{dx}.$ Then the second derivative, i.e. $\delta^2,$ is not a derivation because if it was, then we would have $\delta^2(x^2)=2x\delta^2(x)=0$ but we know that $\delta^2(x^2)=2.$

Example 2. Let $M_n(R)$ be the ring of $n \times n$ matrices with entries from a ring $R.$ Let $\delta \in \text{Der}(R).$ Define the map $\Delta: M_n(R) \to M_n(R)$ as follows: for any $A=[a_{ij}] \in M_n(R),$ where $a_{ij}$ is the $(i,j)$ -entry of $A,$ define $\Delta(A)=[\delta(a_{ij})].$ Then $\Delta \in \text{Der}(M_n(R)).$

Proof. Since $\delta$ is additive, $\Delta$ is clearly additive too. So we only need to show that $\Delta$ satisfies the product rule. Let $A=[a_{ij}], \ B=[b_{ij}] \in M_n(R).$ Let $c_{ij},d_{ij}$ be, respectively, the $(i,j)$ -entry of $AB$ and the $(i,j)$ -entry of $\Delta(AB).$ Then $c_{ij}=\sum_{k=1}^na_{ik}b_{kj}$ and so

$\displaystyle \begin{aligned} d_{ij}=\delta(c_{ij})=\sum_{k=1}^n\delta(a_{ik}b_{kj})=\sum_{k=1}^n(\delta(a_{ik})b_{kj}+a_{ik}\delta(b_{kj}))=\sum_{k=1}^n\delta(a_{ik})b_{kj}+\sum_{k=1}^na_{ik}\delta(b_{kj})\end{aligned},$

which is the $(i,j)$ -entry of $\Delta(A)B+A\Delta(B).$ So $\Delta(AB)= \Delta(A)B+A\Delta(B)$ hence the result. $\ \Box$

Next is to prove a familiar formula from Calculus, i.e. the Leibniz formula for the $n$ th derivative of the product of two elements.

The Leibniz Formula. Let $R$ be a ring, $a,b \in R,$ and $\delta \in \text{Der}(R).$ Then

$\displaystyle \delta^n(ab)=\sum_{k=0}^n\binom{n}{k}\delta^k(a)\delta^{n-k}(b),$

for all integers $n \ge 0.$ Here $\delta^0$ is defined to be the identity map.

Proof. The proof is by induction over $n.$ There is nothing to prove for $n=0.$ Now, assuming that the formula holds for $n,$ we have

$\displaystyle \delta^{n+1}(ab)=\delta(\delta^n(ab))=\delta \left( \sum_{k=0}^n\binom{n}{k}\delta^k(a)\delta^{n-k}(b)\right)=\sum_{k=0}^n\binom{n}{k}\delta(\delta^k(a)\delta^{n-k}(b))$

$\displaystyle =\sum_{k=0}^n\binom{n}{k}(\delta^{k+1}(a)\delta^{n-k}(b)+\delta^k(a)\delta^{n-k+1}(b)), \ \ \ \ \ \ \text{by the product rule}$

$\displaystyle =\sum_{k=0}^n\binom{n}{k}\delta^{k+1}(a)\delta^{n-k}(b)+\sum_{k=0}^n\binom{n}{k}\delta^k(a)\delta^{n-k+1}(b)$

$\displaystyle =\sum_{k=1}^{n+1}\binom{n}{k-1}\delta^k(a)\delta^{n-k+1}(b)+\sum_{k=0}^n\binom{n}{k}\delta^k(a)\delta^{n-k+1}(b)$

$\displaystyle =\sum_{k=1}^{n+1}\binom{n}{k-1}\delta^k(a)\delta^{n-k+1}(b)+\sum_{k=1}^{n+1}\binom{n}{k}\delta^k(a)\delta^{n-k+1}(b)+a\delta^{n+1}(b)$

$\displaystyle =\sum_{k=1}^{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]\delta^k(a)\delta^{n-k+1}(b)+a\delta^{n+1}(b)$

$\displaystyle =\sum_{k=1}^{n+1}\binom{n+1}{k}\delta^k(a)\delta^{n-k+1}(b)+a\delta^{n+1}(b)=\sum_{k=0}^{n+1}\binom{n+1}{k}\delta^k(a)\delta^{n-k+1}(b),$

which proves the formula for $n+1. \ \Box$

Let me end this post with an example which is Remark 1.6.30 in Louis Rowen’s book Ring Theory (volume 1).

Example 3. Let $R$ be a ring, and $\delta \in \text{Der}(R).$ In the first part of this post, we showed that $\ker \delta$ is a subring of $R.$ Now, for any integer $n \ge 0,$ let

$K_n:=\ker \delta^n=\{a \in R: \ \delta^{n+1}(a)=0\}.$

So $\ker \delta =K_0.$ Then $K:=\bigcup_{k=0}^{\infty}K_n$ is a subring of $R$ and $K_mK_n \subseteq K_{m+n},$ for all $m,n.$

Proof. It is clear that $(K_n,+)$ is an abelian group for all $n,$ and $K_0 \subseteq K_1 \subseteq \cdots.$ So $(K,+)$ is an abelian group, and so we only need to show that $K_mK_n \subseteq K_{m+n}$ for all $m,n.$ Let $a \in K_m$ and $b \in K_n.$ We want to show that $ab \in K_{m+n}.$ So $\delta^{m+1}(a)=\delta^{n+1}(b)=0.$ We also have, by Leibniz formula,

$\displaystyle \delta^{m+n+1}(ab)=\sum_{k=0}^{m+n+1}\binom{m+n+1}{k}\delta^k(a)\delta^{m+n+1-k}(b).$

Now, if $k \ge m+1,$ then $\delta^k(a)=0$ and if $k \le m,$ then $m+n+1-k \ge n+1$ and therefore $\delta^{m+n+1-k}(b)=0.$ Thus $\delta^{m+n+1}(ab)=0$ and so $ab \in K_{m+n}. \ \Box$

Derivations on rings; the basics (1)

Posted: March 14, 2024 in Derivation, Noncommutative Ring Theory Notes
Tags: Derivation, inner derivation, polynomial ring

All rings in this post are assumed to have identity and $Z(R)$ denotes the center of a ring $R.$

Those who have the patience to follow my posts have seen derivation on rings several times but they have not seen a post exclusively about the basics on derivations of rings because, strangely, that post didn’t exist until now.

In the polynomial ring $\mathbb{C}[x],$ we have the familiar concept of differentiation with respect to $x,$ i.e. $\delta:= \frac{d}{dx}.$ This is a map form $\mathbb{C}[x] \to \mathbb{C}[x]$ which is additive and satisfies the product rule. Also, $\delta(c)=0$ for all $c \in \mathbb{C}.$ We now extend this concept to rings in general.

Definition 1. Let $R$ be a ring. A derivation of $R$ is any additive map $\delta : R \to R$ that satisfies the product rule: $\delta(ab)=\delta(a)b+a\delta(b),$ for all $a,b \in R.$ We denote by $\text{Der}(R)$ the set of all derivations of $R.$

We can now extend the familiar concept of differentiating polynomials to differentiating polynomials over any ring $R.$ Note that the variable $x$ is assumed to be in the center of $R[x].$

Example 1. Let $R$ be a ring, and let $R[x]$ be the ring of polynomials over $R.$ Define the map $\delta : R[x] \to R[x]$ by

$\displaystyle \delta\left(\sum_{k=0}^na_kx^k\right)=\sum_{k=0}^nka_kx^{k-1}, \ \ \ \ \ n \in \mathbb{Z}_{\ge 0}, \ \ a_k \in R.$

Then $\delta \in \text{Der}(R[x]).$ The derivation $\delta$ is usually denoted by $\frac{d}{dx}.$ Note that, for the definition to make sense, the term $ka_kx^{k-1}$ for $k=0$ is defined to be $0$ and $x^0$ is defined to be $1.$

Proof. We need to show that $\delta$ is additive and also it satisfies the product rule. Let $p(x), q(x) \in R[x].$ So we can write $p(x)=\sum_{k=0}^na_kx^k, \ q(x)=\sum_{k=0}^nb_kx^k.$ Then

$\displaystyle \delta(p(x)+q(x))=\delta \left(\sum_{k=0}^n(a_k+b_k)x^k\right)=\sum_{k=0}^nk(a_k+b_k)x^{k-1}$

$\displaystyle =\sum_{k=0}^nka_kx^{k-1}+\sum_{k=0}^nkb_kx^{k-1}=\delta(p(x))+\delta(q(x)).$

To prove the product rule, we have

$\displaystyle \delta(p(x)q(x))=\delta \left(\sum_{k=0}^{2n}\sum_{j=0}^ka_jb_{k-j}x^k\right)=\sum_{k=0}^{2n}k\sum_{j=0}^ka_jb_{k-j}x^{k-1}, \ \ \ \ \ \ \ \ \ (*)$

and

$\displaystyle x(\delta(p(x))q(x)+p(x)\delta(q(x))=x\left(\sum_{k=0}^nka_kx^{k-1}\sum_{k=0}^nb_kx^k+\sum_{k=0}^na_kx^k\sum_{k=0}^nkb_kx^{k-1}\right)$

$\displaystyle \begin{aligned}=\sum_{k=0}^nka_kx^k\sum_{k=0}^nb_kx^k+\sum_{k=0}^na_kx^k\sum_{k=0}^nkb_kx^k=\sum_{k=0}^{2n}\sum_{j=0}^kja_jb_{k-j}x^k+\sum_{k=0}^{2n}\sum_{j=0}^k(k-j)a_jb_{k-j}x^k\end{aligned}$

$\displaystyle =\sum_{k=0}^{2n}k\sum_{j=0}^ka_jb_{k-j}x^k=x \delta(p(x)q(x)), \ \ \ \ \ \text{by} \ (*).$

Therefore $\delta(p(x)q(x))=\delta(p(x))q(x)+p(x)\delta(q(x)). \ \Box$

Example 2. Let $R$ be a ring, and let $r \in R.$ Define the map $\delta : R \to R$ by $\delta(a)=ra-ar,$ for all $a \in R.$ Then $\delta \in \text{Der}(R)$ and it’s called an inner derivation.

Proof. We need to show that $\delta$ is additive and also it satisfies the product rule. Let $a,b \in R.$ Then

$\delta(a+b)=r(a+b)-(a+b)r=ra-ar+rb-br=\delta(a)+\delta(b),$

and

$\begin{aligned} \delta(a)b+a\delta(b)=(ra-ar)b+a(rb-br)=rab-arb+arb-abr=rab-abr=\delta(ab). \ \Box \end{aligned}$

Remarks. Let $R$ be a ring, and let $\delta \in \text{Der}(R).$ Let $K:=\ker \delta=\{a \in R: \ \delta(a)=0\}.$

i) $\{0,1\} \subseteq K,$

ii) $K$ is a subring of $R$ called the ring of constants of $\delta,$

iii) if $a \in K$ is invertible in $R,$ then $a^{-1} \in K;$ so if $R$ is a division ring, $K$ is a division ring too,

iv) if $r \in R,$ then $\delta(ra)=r\delta(a)$ for all $a \in R$ if and only if $r \in K,$

v) $\delta(Z(R)) \subseteq Z(R),$

vi) if $\delta(a) \in Z(R),$ then $\delta(a^n)=na^{n-1}\delta(a)$ for all integers $n \ge 1,$ where, for $n=1,$ we define $a^0=1,$

vii) The identity given in vi) may not hold true if $\delta(a) \notin Z(R).$

Proof. i) Since $\delta$ is additive, $\delta(0)=\delta(0+0)=\delta(0)+\delta(0)=2\delta(0),$ and so $\delta(0)=0,$ hence $0 \in K.$ Also, by product rule, $\delta(1)=\delta(1 \cdot 1)=\delta(1) \cdot 1 + 1 \cdot \delta(1)=2\delta(1),$ and so $\delta(1)=0$ hence $1 \in K.$

ii) If $a,b \in \ker \delta,$ then since $\delta$ is additive, $\delta(a+b)=\delta(a)+\delta(b)=0+0=0$ and so $a+b \in \ker \delta.$ Also, by product rule, $\delta(ab)=\delta(a)b+a\delta(b)=0+0=0$ and so $ab \in \ker \delta.$

iii) By i) and product rule, $0=\delta(1)=\delta(aa^{-1})=\delta(a)a^{-1}+a\delta(a^{-1})=a\delta(a^{-1})$ and so $\delta(a^{-1}) =0.$

iv) If $\delta(ra)=r\delta(a),$ for some $r \in R$ and all $a \in R,$ then $a=1$ gives $\delta(r)=r\delta(1)=0$ and so $r \in K.$ Conversely, if $r \in K, \ a \in R,$ then, by product rule, $\delta(ra)=\delta(r)a+r\delta(a)=0+r\delta(a)=r\delta(a).$

v) Let $a \in Z(R), \ b \in R.$ Then

$\delta(a)b+a\delta(b)=\delta(ab)=\delta(ba)=\delta(b)a+b\delta(a)=a\delta(b)+b\delta(a),$

and so $\delta(a)b=b\delta(a),$ proving that $\delta(a) \in Z(R).$

vi) First notice that, by v), the condition $\delta(a) \in Z(R)$ is stronger than $a \in Z(R).$ Now, the proof is by induction over $n.$ It is clear for $n=1.$ Assuming the identity holds for $n,$ we have

$\begin{aligned} \delta(a^{n+1})=\delta(a^na)=\delta(a^n)a+a^n\delta(a)=na^{n-1}\delta(a)a+a^n\delta(a)=na^n\delta(a)+a^n\delta(a)=(n+1)a^n\delta(a).\end{aligned}$

vii) Consider $R:=M_2(\mathbb{C}),$ the ring of $2 \times 2$ matrices with complex entries. Let $\delta$ be the inner derivation (see Example 2) of $R$ corresponding to $r=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}.$ Now, choose $a=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}.$ Then $a^2$ is the identity matrix and so $\delta(a^2)=0.$ Also,

$\delta(a)=ra-ar=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \notin Z(R), \ \ \ 2a\delta(a) =\begin{pmatrix}-2 & 0 \\ 0 & 2\end{pmatrix} \ne 0=\delta(a^2). \ \Box$

Definition 2. Let $R$ be a ring, $\delta \in \text{Der}(R),$ and $C$ a subring of $R$ contained in the center of $R.$ If $C \subseteq \ker \delta,$ then, by Remark iv), $\delta(ca+b)=c\delta(a)+\delta(b)$ for all $c \in C,a,b \in R,$ i.e. $\delta$ is $C$ -linear. The set of all $C$ -linear derivations of $R$ is denoted by $\text{Der}_C(R).$

Exercise. Consider the commutative polynomial ring $C]x]$ and let $R:=C[x]/I,$ where $I:=\langle x^2\rangle.$ Define the map $\delta: R \to R$ by $\delta(\alpha x+ \beta+I)=\alpha x + I,$ for all $\alpha, \beta \in C.$ Show that $\delta \in \text{Der}_C(R).$

In part two of this post, we will learn more about derivations on rings. Happy $\pi$ day!

PI rings; definition and basic examples

Posted: February 20, 2024 in Noncommutative Ring Theory Notes, Polynomial Identity Rings
Tags: PI algebra, PI ring, polynomial identity ring, real quaternion algebra

Rings in this post may or may not have identity. As always, the ring of $n \times n$ matrices with entries from a ring $R$ is denoted by $M_n(R).$

PI rings are arguably the most important generalization of commutative rings. This post is the first part of a series of posts about this fascinating class of rings.

Let $C$ be a commutative ring with identity, and let $C\langle x_1, \ldots ,x_n\rangle$ be the ring of polynomials in noncommuting indeterminates $x_1, \ldots ,x_n$ and with coefficients in $C.$ We will assume that each $x_i$ commutes with every element of $C.$ If $n=1,$ then $C\langle x_1, \ldots ,x_n\rangle$ is just the ordinary commutative polynomial ring $C[x].$
A monomial is an element of $C\langle x_1, \ldots ,x_n\rangle$ which is of the form $y_1y_2 \cdots y_k,$ where $y_i \in \{x_1, \cdots , x_n\}$ for all $i.$ The degree of a monomial $y_1y_2 \cdots y_k$ is defined to be $k.$ For example, $x_1x_2x_1^3x_5^2$ is a monomial of degree $7.$ So an element of $f \in C\langle x_1, \ldots ,x_n\rangle$ is a $C$ -linear combination of monomials, and we say that $f$ is monic if the coefficient of at least one of the monomials of the highest degree in $f$ is $1.$ For example, $x_1^2+x_2-3x_2x_3x_2+2x_4^3$ is not monic because none of the monomials of the highest degree, i.e. $x_2x_3x_2$ and $x_4^3,$ have coefficient $1,$ but $x_1^2+x_2-3x_2x_3x_2+x_4^3$ is monic.

Definition 1. A ring $R$ is called a polynomial identity ring, or PI ring for short, if there exists a positive integer $n$ and a monic polynomial $f \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle$ such that $f(r_1, \cdots , r_n)=0$ for all $r_i \in R.$ We then say that $R$ satisfies $f$ or $f$ is an identity of $R.$

Definition 2. Let $C$ be a commutative ring with identity, and let $R$ be a $C$ -algebra. If, in Definition 1, we replace $\mathbb{Z}$ with $C,$ we will get the definition of a PI algebra. So $R$ is called a PI algebra if there exists a positive integer $n$ and a monic polynomial $f \in C\langle x_1, \ldots ,x_n\rangle$ such that $f(r_1, \cdots , r_n)=0$ for all $r_i \in R.$ Note that since every ring is a $\mathbb{Z}$ -algebra, every PI ring is a PI algebra.

Example 1. Every commutative ring $R$ is a PI ring.

Proof. Since $R$ is commutative, $r_1r_2-r_2r_1=0,$ for all $r_1,r_2 \in R,$ and therefore $R$ satisfies the monic polynomial $f=x_1x_2-x_2x_1. \ \Box$

Remark 1. A PI ring could satisfy many polynomials. For example a finite field of order $q$ satisfies the polynomial in Example 1, because it is commutative, and it also satisfies the polynomial $x^q-x.$ Another example is Boolean rings; they satisfy the polynomial in Example 1, because they are commutative, and they also satisfy the polynomial $x^2-x.$

Example 2. If $C$ is a commutative ring with identity, then $R:=M_2(C)$ is a PI ring.

Proof. Let $A_1,A_2 \in R.$ Then $\text{tr}(A_1A_2 - A_2A_1)=0$ and so, by Cayley-Hamilton,

$(A_1A_2 - A_2A_1)^2 = cI_2,$

for some $c \in C.$ Thus $(A_1A_2-A_2A_1)^2$ commutes with every element of $R,$ i.e.,

$(A_1A_2-A_2A_1)^2A_3-A_3(A_1A_2-A_2A_1)^2=0$

for all $A_1,A_2,A_3 \in R.$ Thus $R$ satisfies the monic polynomial

$f=(x_1x_2-x_2x_1)^2x_3 - x_3(x_1x_2 - x_2x_1)^2. \ \Box$

Example 3. The division ring of real quaternions $\mathbb{H}$ is a PI ring.

Proof. Recall that $\mathbb{H}=\mathbb{R}+\mathbb{R}\bold{i}+\mathbb{R}\bold{j}+\mathbb{R}\bold{k},$ where $\bold{i}^2=\bold{j}^2=-1, \ \bold{ij}=-\bold{ji}=\bold{k}.$ Let

$x=a+b\bold{i}+c\bold{j}+d\bold{k} \in \mathbb{H}, \ \ \ \ \ a,b,c,d \in \mathbb{R}.$

It is easy to see that $x^2-2ax+a^2+b^2+c^2+d^2=0.$ Thus, since $a,b,c,d$ are in the center of $\mathbb{H},$ we get that $y(x^2-2ax)=(x^2-2ax)y$ for all $y \in \mathbb{H}.$ So $yx^2-x^2y=2a(yx-xy)$ and hence, since $a$ is central, $(yx-xy)(yx^2-x^2y)=(yx^2-x^2y)(yx-xy)$ for all $x,y \in \mathbb{H}.$ Therefore $\mathbb{H}$ satisfies the monic polynomial $f=(x_1x_2-x_2x_1)(x_1x_2^2-x_2^2x_1)-(x_1x_2^2-x_2^2x_1)(x_1x_2-x_2x_1). \ \Box$

Remark 2. If $R$ is a PI ring with identity, then $R$ could satisfy a polynomial with a nonzero constant. For example, the ring $\mathbb{Z}/n\mathbb{Z}$ satisfies the polynomial in Example 1, and it also satisfies the polynomial $f=(x-1)(x-2) \cdots (x-n)+n.$ Now suppose that $R$ has no identity, and $f \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle$ has a nonzero constant $m.$ So $f=g + m,$ where $g \in \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle$ has a zero constant. Now, what is $f(r_1, \cdots , r_n), \ r_i \in R?$ Well, It is not defined because it is supposed to be $g(r_1, \cdots , r_n)+m1_R$ but $1_R$ does not exist. So if $R$ has no identity, all identities of $R$ must have zero constants.

Example 4. Any subring or homomorphic image of a PI ring is a PI ring.

Proof. If $R$ satisfies $f,$ then obviously any subring of $R$ satisfies $f$ too. If $S$ is a homomorphic image of $R,$ then $S \cong R/I$ for some ideal $I$ of $R.$ Now, it is clear that for any $f(x_1, \cdots , x_n) \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle$ and any $r_1, \cdots , r_n \in R,$ we have $f(r_1+I, \cdots , r_n+I)=f(r_1, \cdots , r_n) + I.$ Hence if $R$ satisfies $f,$ then $R/I$ satisfies $f$ too. $\ \Box$

Example 5. If $I$ is a nilpotent ideal of a ring $R,$ and $R/I$ is a PI ring, then $R$ is a PI ring.

Proof. So $I^m=(0)$ for some positive integer $m.$ Now suppose that $R/I$ satisfies a monic polynomial $f(x_1, \cdots , x_n) \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle.$ Then

$I=0_{R/I}=f(r_1+I, \cdots , r_n+I)=f(r_1, \cdots , r_n) + I$

for all $r_i \in R,$ and so $f(r_1, \cdots , r_n) \in I.$ Thus $(f(r_1, \cdots , r_n))^m \in I^m=(0)$ and hence $R$ satisfies the monic polynomial $f^m. \ \Box$

Example 6. A finite direct product of PI rings is a PI ring.

Proof. By induction, we only need to prove that for a direct product two PI rings. So let $R_1, R_2$ be PI rings that, respectively, satisfy $f,g \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle,$ and let $R:=R_1 \times R_2.$ It is clear that for any polynomial $h \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle$ and $(r_i,s_i) \in R, \ 1 \le i \le n,$ we have

$h((r_1,s_1), \cdots , (r_n,s_n))=(h(r_1, \cdots , r_n),h(s_1, \cdots , s_n)).$

Therefore

$f((r_1,s_1), \cdots , (r_n,s_n))g((r_1,s_1), \cdots , (r_n,s_n))=(f(r_1, \cdots , r_n)g(r_1, \cdots , r_n), f(s_1, \cdots , s_n)g(s_1, \cdots , s_n))$

$=(0_{R_1},0_{R_2})=0_R,$

because $f(r_1, \cdots , r_n)=0_{R_1}$ and $g(s_1, \cdots , s_n)=0_{R_2}.$ So $R$ satisfies the monic polynomial $fg. \ \Box$

Exercise. Let $R$ be a ring with the center $C.$ Suppose that for every $r \in R,$ there exist $a,b,c \in C$ such that $r^3+ar^2+br+c=0.$ Show that $R$ is a PI ring.
Hint. See the proof of Example 3.

Note. The reference for this post is Section 1, Chapter 13 of the book Noncommutative Noetherian Rings by McConnell and Robson. Example 3 was added by me.

Left-multiplication maps as ring homomorphisms

Posted: February 14, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: abelian ring, Artin-Wedderburn theorem, center of rings, idempotent, left-multiplication, prime ring, SE, semiprime ring, semiprimitive ring, semisimple ring, upper triangular matrix

Throughout this post, $R$ is a ring with identity and $Z(R)$ is its center. We denote by $\mathcal{I}(R)$ the set of idempotents of $R,$ i.e. $r \in R$ such that $r^2=r.$ Also, $R[x]$ is the ring of polynomials in $x$ with coefficients in $R.$ Given $r \in R,$ the left-multiplication map $\ell_r: R \to R$ is defined by $\ell_r(a)=ra,$ for all $a \in R.$ Finally, ring homomorphisms in this post are not required to preserve the identity element.

Left multiplication maps $\ell_r$ are clearly right $R$ -module homomorphisms, i.e. $\ell_r(ab)=\ell_r(a)b,$ for all $a,b \in R.$ But for which $r \in R$ is the map $\ell_r$ a ring homomorphism? That’s probably not easy to answer in general, but I’m going to share some of my findings with you anyway. Before I do that, let me define a class of rings with a funny name which are related to our question.

Definition. A ring $R$ is called abelian if $\mathcal{I}(R) \subseteq Z(R),$ i.e. every idempotent of $R$ is central.

Commutative rings are clearly abelian but abelian rings are not necessarily commutative. For example, by Remark 3 in this post, every reduced ring is abelian (in fact, by Exercise 1 in that post, every reversible ring is abelian). Part vi) of the following problem gives an equivalent definition of abelian rings: a ring $R$ is abelian if and only if $\ell_r$ is a ring homomorphism for every idempotent $r$ of $R.$

Problem (Y. Sharifi). Let $R$ be a ring, and consider the set

$\mathcal{L}(R):=\{r \in R: \ \ell_r \ \text{is a ring homomorphism}\}.$

Show that

i) $\mathcal{L}(R)=\{r \in R: \ ra=rar, \ \forall a \in R\},$

ii) if $R$ is commutative, then $\mathcal{L}(R)=\mathcal{I}(R),$

iii) $\mathcal{L}(R)$ is multiplicatively closed,

iv) given $r \in \mathcal{I}(R),$ the set $S_r:=\{a \in R: \ ra=rar\}$ is a subring of $R,$

v) $Z(R) \cap \mathcal{I}(R) \subseteq \mathcal{L}(R) \subseteq \mathcal{I}(R),$

vi) $\mathcal{L}(R) = \mathcal{I}(R)$ if and only if $R$ is abelian,

vii) if $R$ is semiprime, then $\mathcal{L}(R)= Z(R) \cap \mathcal{I}(R),$

viii) if $R$ is prime, then $\mathcal{L}(R)=\{0,1\},$

ix) $\mathcal{L}(R) =R \cap \mathcal{L}(R[x])$ and if $R$ is commutative or semiprime, then $\mathcal{L}(R[x])=\mathcal{L}(R),$

x) $\mathcal{L}\left(\prod_{i \in I}R_i\right)=\prod_{i \in I}\mathcal{L}(R_i)$ for any family of rings $\{R_i\}_{i \in I}.$

Solution. i) Since $\ell_r$ is clearly additive, it is a ring homomorphism if and only if $\ell_r(ab)=\ell_r(a)\ell_r(b),$ i.e. $rab=rarb$ for all $a,b \in R.$ Choosing $b=1$ gives $ra=rar.$ Conversely, if $ra=rar,$ then $rab=rarb$ and so $\ell_r$ is a ring homomorphism.

ii) By i), $r \in \mathcal{L}(R)$ if and only if $ra=r^2a,$ for all $a \in R,$ which holds if and only if $r^2=r,$ i.e $r \in \mathcal{I}(R).$

iii) Let $r,s \in \mathcal{L}(R)$ and $a \in R.$ Then, by i), $rsar=rsa, \ sas=sa,$ and so $rsars=rsas=rsa.$ Thus $rs \in \mathcal{L}(R).$

iv) If $a,b \in S_r,$ then $rab=rarb=rarbr=rabr$ and so $ab \in S_r.$

v) Let $r \in \mathcal{L}(R).$ Then choosing $a=1$ in i) gives $r=r^2$ and so $r \in \mathcal{I}(R).$ Now, let $r \in Z(R) \cap \mathcal{I}(R),$ and $a \in R.$ Then $rar=rra=r^2a=ra$ and so $r \in \mathcal{L}(R).$

vi) If $\mathcal{I}(R) \subseteq Z(R),$ then $\mathcal{L}(R) = \mathcal{I}(R),$ by v). Suppose now that $\mathcal{L}(R) = \mathcal{I}(R)$ and $r \in \mathcal{I}(R).$ Then $1-r \in \mathcal{I}(R),$ and so, by i), $ra=rar, \ (1-r)a=(1-r)a(1-r),$ for all $a \in R.$ Thus

$a-ra=(1-r)a=(1-r)a(1-r)=a-ar-ra+rar=a-ar,$

which gives $ra=ar$ and so $r \in Z(R).$

vii) By v), we only need to show that $\mathcal{L}(R) \subseteq Z(R).$ So let $r \in \mathcal{L}(R)$ and $a,b \in R.$ If we show that $(ra-ar)b(ra-ar)=0,$ we are done because that means $(ra-ar)R(ra-ar)=(0)$ and so, since $R$ is semiprime, $ra=ar$ hence $r \in Z(R).$ Now, to show that $(ra-ar)b(ra-ar)=0,$ we have

$(ra-ar)b(ra-ar)=rabra-rabar-arbra+arbar$

and so, since by i), $rabr=rab, \ rbr=rb, \ rbar=rba,$ we get that

$(ra-ar)b(ra-ar)=raba-raba-arba+arba=0.$

viii) Clearly the inclusion $\{0,1\} \subseteq \mathcal{L}(R)$ holds for any ring $R.$ Now, suppose that $R$ is a prime ring and $r \in \mathcal{L}(R), a \in R.$ Then, by i), $ra(1-r)=0$ and so $rR(1-r)=(0),$ which gives $r=0$ or $r=1.$

ix) Let $r \in R.$ Then, by i), $r \in \mathcal{L}(R[x])$ if and only if $rf(x)=rf(x)r$ for all $f(x) \in R[x]$ if and only if $ra=rar$ for all $a \in R,$ because $x$ is central in $R[x],$ if and only if $r \in \mathcal{L}(R),$ by i).
If $R$ is commutative, then $\mathcal{L}(R)=\mathcal{I}(R)$ and so, since the polynomial ring $R[x]$ is commutative too,

$\mathcal{L}(R[x])=\mathcal{I}(R[x])=\mathcal{I}(R)=\mathcal{L}(R).$

Note that the identity $\mathcal{I}(R[x])=\mathcal{I}(R)$ is true by this post.
Now suppose that $R$ is semiprime, and $f(x)=\sum_{i=0}^nr_ix^i \in \mathcal{L}(R[x]), \ n \ge 1.$ So $f(x)a=f(x)af(x)$ for all $a \in R$ and hence, equating the coefficients of $x^{2n}$ on both sides of the equality gives $0=r_nar_n.$ Thus $r_nRr_n=(0)$ and so, since $R$ is semiprime, $r_n=0,$ contradiction. This proves that $n=0$ and hence $f(x) \in R$ giving $\mathcal{L}(R[x]) = \mathcal{L}(R).$

x) Clear, by i). $\ \Box$

Example 1. This example shows that $\mathcal{L}(R)$ is not always central. Let $C$ be a commutative domain, and let $R$ be the ring of $2 \times 2$ upper triangular matrices with entries in $C.$ Then, using the first part of the Problem, it is easy to see that

$\mathcal{L}(R)=\left \{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}0 & c \\ 0 & 1\end{pmatrix}, \ c \in C\right \}.$

Example 2. Let $R$ be the ring in Example 1. Then $\mathcal{L}(R[x]) \neq \mathcal{L}(R)$ because $f(x)=r+sx \in \mathcal{L}(R[x]),$ where

$r=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}, \ \ \ \ \ s=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$

Example 3. If $R$ is a semisimple ring, then $|\mathcal{L}(R)|=2^n$ for some positive integer $n.$

Proof. By the Artin-Wedderburn theorem, $R$ is a finite direct product of matrix rings over division rings. The result now follows from parts viii) and x) of the Problem. $\ \Box$

Semiprime rings

Posted: February 12, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: matrix ring, nilpotent ideal, prime ring, reduced ring, semiprime ideal, semiprime ring, semiprimitive ring

Throughout this post, $R$ is a ring with identity, and $M_n(R)$ is the ring of $n \times n$ matrices with entries from $R.$ Also, by ideal we always mean two-sided ideal.

Here we defined a prime ideal of $R$ as a proper ideal $P$ which satisfies this condition: if $IJ \subseteq P$ for some ideals $I,J$ of $R,$ then $I \subseteq P$ or $J \subseteq P.$ By weakening this condition, we get a set of ideals of $R$ that contains the set of prime ideals of $R;$ we call these ideals semiprime.

Definition 1. A proper ideal $Q$ of a ring $R$ is called semiprime if for any ideal $I$ of $R, \ I^2 \subseteq Q$ implies that $I \subseteq Q.$

Now, what exactly are semiprime ideals in commutative rings?

Remark 1. In a commutative ring $R,$ semiprime ideals are just radical ideals.

Proof. Since $R$ is commutative, the condition $I^2 \subseteq Q \implies I \subseteq Q$ for all ideals $I$ of $R,$ which is the definition of a semiprime ideal $Q,$ is equivalent to the condition $a^2 \in Q \implies a \in Q$ for all $a \in R,$ which itself is equivalent to $a^n \in Q \implies a \in Q$ for all $a \in R$ and all positive integers $n$ (why?). Now, the set

$\{a \in R: \ a^n \in Q, \ \text{for some positive integer} \ n\}$

is, by definition, $\sqrt{Q},$ the radical of $Q,$ which is easily seen to be an ideal containing $Q.$ If $\sqrt{Q}=Q$ or, equivalently, $\sqrt{Q} \subseteq Q,$ then we say that $Q$ is a radical ideal. So semiprime ideals of a commutative ring are just radical ideals of the ring. $\ \Box$

Remark 2. It is clear from Definition 1 that any intersection of prime ideals is semiprime. We also know from commutative ring theory that radical ideals are exactly those ideals that are intersection of some prime ideals. By Remark 1, radical ideals of a commutative ring are exactly semiprime ideals of the ring. So it is natural to ask if, in general ring theory, it is true that every semiprime ideal is an intersection of some prime ideals. It turns out that the answer is positive. In fact some authors define a semiprime ideal as an ideal which is an intersection of some prime ideals and then they show this is equivalent to Definition 1. But I did not choose this approach here because it will complicate the proof of the following Proposition hence getting into the goal of this post which is introducing semiprime rings.

By Remark 1, an ideal $Q$ in a commutative ring $R$ is semiprime if and only if $a^2 \in Q \implies a \in Q$ for all $a \in R.$ In general ring theory, we have the following similar result (compare with Proposition 1 in this post!).

Proposition. A proper ideal $Q$ of a ring $R$ is semiprime if and only if for any $a \in R, \ aRa \subseteq Q$ implies that $a \in Q.$

Proof. Suppose first that $Q$ is semiprime and $aRa \subseteq Q$ for some $a \in R.$ Let $I:=RaR.$ Then $I$ is an ideal of $R$ and $I^2=RaRaR=R(aRa)R \subseteq Q.$ Thus $I \subseteq Q,$ which gives $a \in Q.$
Conversely, suppose now that $I^2 \subseteq Q$ for some ideal $I$ of $R$ and $a \in I.$ Then

$aRa \subseteq R(aRa)R=(RaR)(RaR) \subseteq I^2 \subseteq Q$

and so $a \in Q.$ Hence $I \subseteq Q$ which proves that $Q$ is semiprime. $\ \Box$

By Remark 1, $(0)$ is a semiprime ideal of a commutative ring $R$ if and only if $\sqrt{(0)}=(0),$ i.e. if $a^n=0$ for some $a \in R$ and some positive integer $n,$ then $a=0.$ In other words, $(0)$ is a semiprime ideal of $R$ if and only if $R$ is reduced. In general ring theory, it is true that if $R$ is reduced, then $(0)$ is a semiprime ideal of $R$ (Example 1). However the converse is not true (Example 2). If $(0)$ is a semiprime ideal of a ring, then we will call the ring semiprime. So, in general ring theory, every reduced ring is semiprime but not every semiprime ring is reduced.

Definition 2. We say that a ring $R$ is semiprime if $(0)$ is a semiprime ideal of $R,$ i.e. for any ideal $I$ of $R,$ if $I^2=(0),$ then $I=(0).$

The Proposition gives the following simple yet useful test to check if a ring is semiprime or not.

Corollary. A ring $R$ is semiprime if and only if for any $a \in R, \ aRa=(0)$ implies that $a=0.$

Example 1. Every reduced ring $R$ is semiprime.

Proof. If $aRa=(0),$ for some $a \in R,$ then $a^2=0$ and so $a=0,$ because $R$ is reduced. $\ \Box$

Example 2. Every prime ring is obviously semiprime and so the matrix ring $M_n(k)$ over a field $k$ is a semiprime ring which is not reduced.

Example 3. Every semiprimitive ring $R$ is semiprime.

Proof. Suppose that $aRa=(0),$ for some $a \in R.$ Then $(ra)^2=0$ for all $r \in R$ and hence $1-ra$ is invertible. Thus $a \in J(R)=(0). \ \Box$

Example 4. Let $R$ be a ring and $S:=M_n(R).$ Then $S$ is semiprime if and only if $R$ is semiprime. In particular, if $R$ is reduced, then $S$ is semiprime.

Proof. It’s the same as the proof for prime rings (Example 4). $\ \Box$

Exercise 1. Show that a proper ideal $Q$ of a ring $R$ is semiprime if and only if the ring $R/Q$ is semiprime.

Exercise 2. Show that a proper ideal $Q$ of a ring $R$ is semiprime if and only if $I^2 \subseteq Q$ implies that $I \subseteq Q$ for any left ideal $I$ of $R.$

Exercise 3. Show that the center of a semiprime ring $R$ is reduced.
Hint. If $a^2=0$ for some central element of $R,$ then $aRa=Ra^2=(0).$

Exercise 4. Show that a direct product of rings is semiprime if and only if each ring is semiprime.

Exercise 5. Show that a ring $R$ is semiprime if and only if $R$ has no nonzero nilpotent ideal, i.e if $I^n=(0)$ for some ideal $I$ of $R$ and some positive integer $n,$ then $I=(0).$
Hint. If $n \ge 2$ is the smallest positive integer $n$ such that $I^n=0,$ then $(I^{n-1})^2=I^{2n-2}=(0).$

Note. The references for this post are Chapter 4 of T. Y. Lam’s book A First Course in Noncommutative Rings and Chapter 3 of Goodearl & Warfield’s book An Introduction to Noncommutative Noetherian Rings.

Prime rings

Posted: February 10, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: matrix ring, prime ideal, prime ring, primitive ring, reduced ring, simple ring

Throughout this post, $R$ is a ring with identity, and $M_n(R)$ is the ring of $n \times n$ matrices with entries from $R.$ Also, by ideal we always mean two-sided ideal.

In commutative ring theory, a proper ideal $P$ of a commutative ring $R$ is said to be prime if $IJ \subseteq P,$ where $I,J$ are ideals of $R,$ implies that $I \subseteq P$ or $J \subseteq P.$ The definition of a prime ideal remains the same even if $R$ is not commutative.

Definition 1. A proper ideal $P$ of a ring $R$ is called prime if for any ideals $I,J$ of $R, \ IJ \subseteq P$ implies that $I \subseteq P$ or $J \subseteq P.$

In commutative ring theory, we see that a proper ideal $P$ is prime if and only if $a,b \in R, \ ab \in P$ implies that $a \in P$ or $b \in P.$ In general ring theory, we have the following similar result.

Proposition 1. A proper ideal $P$ of a ring $R$ is prime if and only if for any $a,b \in R, \ aRb \subseteq P$ implies that $a \in P$ or $b \in P.$

Proof. Suppose first that $P$ is prime and $aRb \subseteq P$ for some $a,b \in R.$ Let $I:=RaR, \ J:=RbR.$ Then $I,J$ are ideals of $R$ and $IJ=RaRbR=R(aRb)R \subseteq P.$ Thus either $I \subseteq P,$ which gives $a \in P,$ or $J \subseteq P,$ which gives $b \in P.$
Conversely, suppose now that $IJ \subseteq P$ for some ideals $I,J$ of $R$ and $I \nsubseteq P.$ We need to show that $J \subseteq P.$ Let $a \in I \setminus P, \ b \in J.$ Then $aRb \subseteq R(aRb)R=(RaR)(RbR) \subseteq IJ \subseteq P$ and so $b \in P$ because $a \notin P.$ Hence $J \subseteq P$ which proves that $P$ is prime. $\ \Box$

In commutative ring theory, if $(0)$ is a prime ideal of a ring, the ring is called a commutative domain. In general ring theory, we still call a ring $R$ a domain if $a,b \in R, \ ab=0$ implies that $a=0$ or $b=0.$ It is true that if $R$ is a domain, then $(0)$ is a prime ideal of $R$ (Example 1). However, in general, the converse is not true (see Example 2 and Proposition 2). If $(0)$ is a prime ideal of a ring, then we will call the ring prime. So, in general ring theory, every domain is a prime ring but not every prime ring is a domain.

Definition 2. We say that a ring $R$ is prime if $(0)$ is a prime ideal of $R,$ i.e. for any ideals $I,J$ of $R,$ if $IJ=(0),$ then $I=(0)$ or $J=(0).$

Proposition 1 gives the following simple yet useful test to check if a ring is prime or not. This is useful because it’s in terms of elements not ideals, which are usually not easy to find.

Corollary. A ring $R$ is prime if and only if for any $a,b \in R, \ aRb=(0)$ implies that $a=0$ or $b=0.$

Example 1. Every domain $R$ is prime.

Proof. Suppose that $aRb=(0)$ for some $a,b \in R.$ Then $ab=0$ and so either $a=0$ or $b=0,$ because $R$ is a domain. $\ \Box$

Example 2. Every simple ring $R$ is prime. In particular, every matrix ring $M_n(k)$ over a field $k$ is prime.

Proof. The only ideals of $R$ are $(0), R.$ So $IJ=(0),$ where $I,J$ are ideals of $R,$ implies that $I=(0)$ or $J=(0). \ \Box$

So not every prime ring is a domain. The following characterizes all prime rings that are domain.

Proposition 2. A ring is a domain if and only if it’s both prime and reduced.

Proof. See Remark 2 in this post. $\ \Box$

The following example generalizes Example 2.

Example 3. Every left primitive ring is prime.

Proof. See Fact 2 in this post. $\ \Box$

Example 4. Let $R$ be a ring and $S:=M_n(R).$ Then $S$ is prime if and only if $R$ is prime. In particular, if $R$ is a domain, then $S$ is prime.

Proof. It is a well-known fact that ideals of $S$ are in the form $M_n(I),$ where $I$ is any ideal of $R.$ Now, suppose that $R$ is prime and $KL=(0)$ for some ideals $K,L$ of $S.$ We have $K=M_n(I), \ L=M_n(J)$ for some ideals $I,J$ of $R.$ So $M_n(I)M_n(J)=(0)$ and hence $IJ=(0),$ which gives $I=(0)$ or $J=(0),$ because $R$ is prime. Therefore $K=(0)$ or $L=(0)$ and so $S$ is prime. The proof for the converse is similar. $\ \Box$

Exercise 1. Show that a proper ideal $P$ of a ring $R$ is prime if and only if the ring $R/P$ is prime.

Exercise 2. Show that a proper ideal $P$ of a ring $R$ is prime if and only if $IJ \subseteq P$ implies that $I \subseteq P$ or $J \subseteq P,$ for any left ideals $I,J$ of $R.$

Exercise 3. Show that the center of a prime ring $R$ is a commutative domain.
Hint. If $ab=0$ for some central elements of $R,$ then $aRb=Rab=(0).$

Exercise 4. Show that a direct product of two or more rings can never be a prime ring.
Hint. If $R=R_1 \times R_2,$ and $a=(1,0), \ b=(0,1),$ then $aRb=(0).$

Exercise 5. Show that every maximal ideal of a ring is prime.

Note. The reference for this post is the first few pages of Chapter 4 of T. Y. Lam’s book A First Course in Noncommutative Rings.

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