Archive for the ‘Prime & Semiprime Rings’ Category

Composition of derivations in prime rings; a theorem of Posner

Posted: March 16, 2024 in Derivation, Noncommutative Ring Theory Notes, Prime & Semiprime Rings
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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.

Left-multiplication maps as ring homomorphisms

Posted: February 14, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
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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
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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
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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.