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.

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