As usual, I’ll assume that k is a field. Recall that if a k-algebra A is an Ore domain, then we can localize A at S:=A \setminus \{0\} and get the division algebra Q(A):=S^{-1}A. The algebra Q(A) is called the quotient division algebra of A.

Theorem (Borho and Kraft, 1976) Let A be a finitely generated k-algebra which is a domain of finite GK dimension. Let B be a k-subalgebra of A and suppose that {\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1. Let S:=B \setminus \{0\}. Then S is an Ore subset of A and S^{-1}A=Q(A). Also, Q(A) is finite dimensional as a (left or right) vector space over Q(B).

Proof. First note that, by the corollary in this post, A is an Ore domain and hence both Q(A) and Q(B) exist and they are division algebras. Now, suppose, to the contrary, that S is not (left) Ore. Then there exist x \in S and y \in A such that Sy \cap Ax = \emptyset. This implies that the sum By + Byx + \ldots + Byx^m is direct for any integer m. Let W be a frame of a finitely generated subalgebra B' of B. Let V=W+kx+ky and suppose that A' is the subalgebra of A which is generated by V. For any positive integer n we have

V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}

and thus \dim_k V^{2n} \geq n \dim_k W^n because the sum is direct. So \log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n and hence {\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B'). Taking supremum of both sides over all finitely generated subalgebras B' of B will give us the contradiction {\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B). A similar argument shows that S is right Ore. So we have proved that S is an Ore subset of A. Before we show that S^{-1}A=Q(A), we will prove that Q(B)A=S^{-1}A is finite dimensional as a (left) vector space over Q(B). So let V be a frame of A. For any positive ineteger n, let r(n) = \dim_{Q(B)} Q(B)V^n. Clearly Q(B)V^n \subseteq Q(B)V^{n+1} for all n and

\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A

because \bigcup_{n=0}^{\infty}V^n=A. So we have two possibilities: either Q(B)V^n=Q(B)A for some n or the sequence \{r(n)\} is strictly increasing. If Q(B)V^n = Q(B)A, then we are done because V^n is finite dimensional over k and hence Q(B)V^n is finite dimensional over Q(B). Now suppose that the sequence \{r(n)\} is strictly increasing. Then r(n) > n because r(0)=\dim_{Q(B)}Q(B)=1. Fix an integer n and let e_1, \ldots , e_{r(n)} be a Q(B)-basis for Q(B)V^n. Clearly we may assume that e_i \in V^n for all i. Let W be a frame of a finitely generated subalgebra of B. Then

(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},

which gives us

\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,

because the sum W^ne_1 + \ldots + W^ne_{r(n)} is direct. Therefore {\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B), which is a contradiction. So we have proved that the second possibility is in fact impossible and hence Q(B)A is finite dimensional over Q(B). Finally, since, as we just proved, \dim_{Q(B)}Q(B)A < \infty, the domain Q(B)A is algebraic over Q(B) and thus it is a division algebra. Hence Q(B)A=Q(A) because A \subseteq Q(B)A \subseteq Q(A) and Q(A) is the smallest division algebra containing A. \Box

Throughout R is a ring with 1 and all modules are left R-modules. In Definition 2 in this post, we defined Z(M), the singular submodule of a module M.

Problem 1. Let M be an R-module and suppose that N_1, \cdots, N_k are submodules of M. Prove that \bigcap_{i=1}^k N_i \subseteq_e M if and only if N_i \subseteq_e M for all i.

Solution. We only need to solve the problem for k = 2. If N_1 \cap N_2 \subseteq_e M, then N_1 \subseteq_e M and N_2 \subseteq_e M because both N_1 and N_2 contain N_1 \cap N_2. Conversely, let P be a nonzero submodule of M. Then N_1 \cap P \neq \{0\} because N_1 \subseteq_e M and therefore (N_1 \cap N_2) \cap P = N_2 \cap (N_1 \cap P) \neq \{0\} because N_2 \subseteq_e M. \ \Box

Problem 2. Prove that if M is an R-module, then Z(M) is a submodule of M and Z(R) is a proper two-sided ideal of R. In particular, if R is a simple ring, then Z(R)=\{0\}.

Solution. First note that 0 \in Z(M) because \text{ann}(0)=R \subseteq_e R. Now suppose that x_1,x_2 \in Z(M). Then \text{ann}(x_1+x_2) \supseteq \text{ann}(x_1) \cap \text{ann}(x_2) \subseteq_e M, by Problem 1. Therefore \text{ann}(x_1+x_2) \subseteq_e M and hence x_1+x_2 \in Z(M). Now let r \in R and x \in Z(M). We need to show that rx \in Z(M). Let J be a nonzero left ideal of R. Then Jr is also a left ideal of R. If Jr = \{0\}, then J \subseteq \text{ann}(rx) and thus \text{ann}(rx) \cap J = J \neq \{0 \}. If Jr \neq \{0\}, then \text{ann}(x) \cap Jr \neq \{0\} because x \in Z(M). So there exists s \in J such that sr \neq 0 and srx = 0. Hence 0 \neq s \in \text{ann}(rx) \cap J. So rx \in Z(M) and thus Z(M) is a submodule of M. Now, considering R as a left R-module, Z(R) is a left ideal of R, by what we have just proved. To see why Z(M) is a right ideal, let r \in R and x \in Z(R). Then \text{ann}(xr) \supseteq \text{ann}(x) \subseteq_e R and so \text{ann}(xr) \subseteq_e R, i.e. xr \in Z(R). Finally, Z(R) is proper because \text{ann}(1)=\{0\} and so 1 \notin Z(R). \ \Box

Problem 3. Prove that if M_i, \ i \in I, are R-modules, then Z(\bigoplus_{i \in I} M_i) = \bigoplus_{i \in I} Z(M_i). Conclude that if R is a semisimple ring, then Z(R)=\{0\}.

Solution. The first part is a trivial result of Problem 1 and this fact that if x = x_1 + \cdots + x_n, where the sum is direct, then \text{ann}(x) = \bigcap_{i=1}^n \text{ann}(x_i). The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem. \Box

Problem 4. Suppose that R is commutative and let N(R) be the nilradical of R. Prove that

1) N(R) \subseteq Z(R);

2) it is possible to have N(R) \neq Z(R);

3) if Z(R) \neq \{0\}, then N(R) \subseteq_e Z(R), as R-modules or Z(R)-modules.

Solution. 1) Let a \in N(R). Then a^n = 0 for some integer n \geq 1. Now suppose that 0 \neq r \in R. Then ra^n=0. Let m \geq 1 be the smallest integer such that ra^m = 0. Then 0 \neq ra^{m-1} \in \text{ann}(a) \cap Rr and hence a \in Z(R).

2) Let R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1 and put R=\prod_{i=1}^{\infty}R_i. For every i, let a_i = 2 + 2^i \mathbb{Z} and consider a = (a_1,a_2, \cdots ) \in R. It is easy to see that a \in Z(R) \setminus N(R).

3) Let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra \neq \{0\} and thus there exists r \in R such that ra \neq 0 and ra^2=0. Hence (ra)^2 = 0 and so ra \in N(R). Thus 0 \neq ra \in N(R) \cap Ra implying that N(R) is an essential R-submodule of Z(R). Now, we view Z(R) as a ring and we want to prove that N(R) as an essential ideal of Z(R). Again,  let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra^2 \neq \{0\} and thus there exists r \in R such that ra^2 \neq 0 and ra^3 = 0. Let s = ra \in Z(R). Then (sa)^2=0 and thus 0 \neq sa \in N(R) \cap Z(R)a implying that N(R) is an essential ideal of Z(R). \ \Box

We will assume that R is a ring (not necessarily commutative) with 1 and all modules are left R-modules.

Definition 1. Let M be an R-module and N a nonzero submodule of M. We say that N is an essential submodule of M, and we will write N \subseteq_e M, if N \cap X \neq (0) for any nonzero submodule X of M. Clearly, that is equivalent to saying N \cap Rx \neq (0) for any nonzero element x \in M. So, in particular, a nonzero left ideal I of R is an essential left ideal of R if I \cap J \neq (0) for any nonzero left ideal J of R, which is equivalent to the condition I \cap Rr \neq (0) for any nonzero element r \in R.

Definition 2. Let M be an R-module and x \in M. Recall that the (left) annihilator of x in R is defined by \text{ann}(x)=\{r \in R: \ rx = 0 \}, which is obviously a left ideal of R. Now, consider the set Z(M):=\{x \in M: \ \text{ann}(x) \subseteq_e R \}. It is easy to see that Z(M) is a submodule of M (see Problem 2 in this post for the proof!) and we will call it the singular submodule of M. If Z(M)=M, then M is called singular. If Z(M)=(0), then M is called nonsingular. We will not discuss nonsingular modules in this post.

Problem 1. Prove that if F is a free R-module, then Z(F) \neq F, i.e. a free module is never singular.

Solution. Let x \in F be any element of an R-basis of F. Let r \in \text{ann}(x). Then rx = 0 and so r = 0. Thus \text{ann}(x)=(0) and so x \notin Z(F). \ \Box

Next problem characterizes singular modules.

Problem 2. Prove that an R-module M is singular if and only if M = A/B for some R-module A and some submodule B \subseteq_e A.

Solution. Suppose first that M=A/B where A is an R-module and B \subseteq_e A. Let x = a + B \in M and let J be a nonzero left ideal of R. If Ja = (0), then Ja \subseteq B and so \text{ann}(x) \cap J = J \neq (0). If Ja \neq (0), then B \cap Ja \neq (0) because B \subseteq_e A. So there exists r \in J such that 0 \neq ra \in B. That means 0 \neq r \in \text{ann}(x) \cap J. So we have proved that x \in Z(M) and hence Z(M)=M, i.e. M is singular. Conversely, suppose that M is singular. We know that every R-module is the homomorphic image of some free R-module. So there exists a free R-module F and a submodule K of F such that M \cong F/K. So we only need to show that K \subseteq_e F. Note that K \neq (0), by Problem 1. Let \{x_i \} be an R-basis for F and suppose that 0 \neq x \in F. We need to show that there exists s \in R such that 0 \neq sx \in K. We can write, after renaming the indices if necessarily,

x = \sum_{i=1}^n r_ix_i, \ \ \ \ \ \ \ \ \ \ (1)

where r_1 \neq 0. For any y \in F, let \overline{y}=y+K \in F/K. Now, since F/K is singular, \text{ann}(\overline{x_1}) \subseteq_e R and so \text{ann}(\overline{x_1}) \cap Rr_1 \neq (0). So there exists s_1 \in R such that s_1r_1 \neq 0 and s_1r_1x_1 \in K. Hence (1) gives us

s_1x= s_1r_1x_1 +\sum_{i=2}^n s_1r_ix_i. \ \ \ \ \ \ \ \ \ (2)

Note that s_1x \neq 0 because s_1r_1 \neq 0. Now, if s_1r_i = 0 for all 2 \leq i \leq n, then s_1x =s_1r_1x_1 \in K and we are done. Otherwise, after renaming the indices in the sum on the right hand side of (2) if necessary, we may assume that s_1r_2 \neq 0. Repeating the above process gives us some s_2 \in R such that s_2s_1r_2 \neq 0 and s_2s_1r_2x_2 \in K. Then (2) implies

s_2s_1x = s_2s_1r_1x_1 + s_2s_1r_2x_2 + \sum_{i=3}^n s_2s_1r_ix_i. \ \ \ \ \ \ \ (3)

The first two terms on the right hand side of (3) are in K and s_2s_1x \neq0 because s_2s_1r_2 \neq 0. If we continue this process, we will eventually have a positive integer 1 \leq m \leq n and s = s_ms_{m-1} \cdots s_1 \in R such that 0 \neq sx \in K. \ \Box

Problem. Let F be a field and suppose that f(x),g(x), p(x) are three polynomials in F[x]. Prove that if both f(x) and p(x) are irreducible and p(x) \mid f(g(x)), then \deg f(x) \mid \deg p(x).

Solution. Let \mathfrak{m} and \mathfrak{n} be the ideals of F[x] generated by f(x) and p(x), respectively. Let E=F[x]/\mathfrak{m} and L = F[x]/\mathfrak{n}. Since both f(x) and p(x) are irreducible, E and L are field extensions of F. Now, define the map \varphi : E \longrightarrow L by \varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n}, for all h(x) \in F[x]. We first show that \varphi is well-defined. To see this, suppose that h(x) \in \mathfrak{m}. Then h(x)=f(x)u(x) for some u(x) \in F[x] and hence h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n}, because p(x) \mid f(g(x)). So \varphi is well-defined. Now \varphi is clearly a ring homomorphism and, since E is a field and \ker \varphi is an ideal of E, we must have \ker \varphi = \{0\}. Therefore we may assume that F \subseteq E \subseteq L and hence \deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box

We will assume that R is a commutative ring with 1. We will allow the case R=(0), i.e. where 1_R=0_R. The goal is to describe  the Krull dimension of R in terms of elements of R and not prime ideals of R. The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the Krull dimension of R is the largest integer n \geq 0 for which there exist prime ideals P_i, \ 0 \leq i \leq n, of R such that P_0 \subset P_1 \subset \ldots \subset P_n. Then we write \text{K.dim}(R)=n. If there is no such integer, then we define \text{K.dim}(R)=\infty and if R=(0), we define \text{K.dim}(R)=-1.

Notation. Let x \in R and set S_x = \{x^n(1+ax): \ n \geq 0, \ a \in R \}. Clearly S_x is multiplicatively closed. Let R_x be the localization of R at S_x. If 0 \in S_x, then we define R_x =(0).

Problem 1. Let x \in R and suppose that \mathfrak{m} is a maximal ideal of R. Then \mathfrak{m} \cap S_x \neq \emptyset. Moreover, if P \subset \mathfrak{m} is a prime ideal and x \in \mathfrak{m} \setminus P, then P \cap S_x = \emptyset.

Solution. if x \in \mathfrak{m}, then x \in \mathfrak{m} \cap S_x and we are done. Otherwise, \mathfrak{m} + Rx = R and thus 1 + ax \in \mathfrak{m} for some a \in R. Then 1+ax \in \mathfrak{m} \cap S_x. For the second part, suppose to the contrary that x^n(1+ax) \in P for some integer n \geq 0 and a \in R. Then, since x \notin P, we have 1+ax \in P \subset \mathfrak{m} and thus 1 \in \mathfrak{m} because x \in \mathfrak{m}. \Box

Problem 2. \text{K.dim}(R)=0 if and only if R_x=\{0\} for all x \in R.

Solution. As we already mentioned, R_x = \{0\} means 0 \in S_x. Suppose that 0 \notin S_x for some x \in R. Then there exists a prime ideal P of R such that P \cap S_x = \emptyset, because S_x is multiplicatively closed. By Problem 1, P is not a maximal ideal and thus \text{K.dim}(R) > 0. Conversely, suppose that 0 \in S_x for all x \in R and P is a prime ideal of R. Let \mathfrak{m} be  a maximal ideal of R which contains P. Choose x \in \mathfrak{m} \setminus P. By Problem 1, P \cap S_x =\emptyset, contradicting 0 \in P \cap S_x. \ \Box

Problem 3. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if \text{K.dim}(R_x) \leq n-1 for all x \in R.

Solution. The case n=0 was done in Problem 2. So we’ll assume that n \geq 1. Suppose that \text{K.dim}(R) \leq n and let x \in R. A prime ideal of R_x is in the form PR_x, where P is a prime ideal of R and P \cap S_x = \emptyset. Also, if PR_x and QR_x are two prime ideals of R_x with PR_x \subset QR_x, then P \subset Q. Now, suppose to the contrary that \text{K.dim}(R_x) \geq n and consider the chain P_0R_x \subset P_1R_x \subset \ldots \subset P_n R_x of prime ideals of R_x. By Problem 1, P_n is not a maximal ideal and so there exists a prime ideal P_{n+1} such that P_n \subset P_{n+1} and that will give us the chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R, contradicting \text{K.dim}(R) \leq n. Conversely, suppose that \text{K.dim}(R_x) \leq n-1 for all x \in R. Suppose also, to the contrary, that there exists a chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R. Let x \in P_{n+1} \setminus P_n. By the second part of Problem 1, P_n \cap S_x = \emptyset and thus P_0R_x \subset P_1R_x \subset \ldots \subset P_nR_x is a chain of prime ideals of R_x, contradicting \text{K.dim}(R_x) \leq n-1. \ \Box

Problem 4. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if for every x_0, x_1, \ldots , x_n \in R there exist integers k_0, k_1, \ldots , k_n \geq 0 and a_0, a_1, \ldots ,a_n \in R such that

x_n^{k_n}( \ldots (x_2^{k_2}(x_1^{k_1}(x_0^{k_0}(1+a_0x_0)+a_1x_1) + a_2x_2) + \ldots) + a_nx_n)=0.

Solution. By Problem 3, \text{K.dim}(R) \leq n if and only if \text{K.dim}((\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n})=-1. Therefore \text{K.dim}(R) \leq n if and only if (\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box

Throughout G is a group with the center Z(G) and the commutator subgroup G'. The goal is to prove that if G/Z(G) is finite, then G' is finite too. We will also find an upper bound for |G'| in terms of |G/Z(G)|.

Notation. For any a,b \in G, we define [a,b]=aba^{-1}b^{-1} and a^b = bab^{-1}.

Problem 1. Let a,b,c \in G.

1) [a,b]=b^ab^{-1}, \ a^a=a, \ ba=a^bb and [a,b]^c=[a^c,b^c].

2) c[a,b]=[a^c,b^c]c.

3) If [G:Z(G)]=n, then [a,b]^{n+1}=[a,b^2][a^b,b]^{n-1}.

Proof. 1) The first three identities are trivial and the fourth one is true because the map f:G \longrightarrow G defined by f(g)=cgc^{-1}=g^c is a group homomorphism.

2) By 1) we have [a^c,b^c]c =[a,b]^cc=c[a,b].

3) So g^n \in Z(G) for all g \in G, because [G:Z(G)]=n. So [a,b]^nb^{-1}=b^{-1}[a,b]^n with 1) give us

[a,b]^{n+1}=[a,b][a,b]^n =b^ab^{-1}[a,b]^n = b^a[a,b]^n b^{-1}=b^a[a,b][a,b]^{n-1}b^{-1}

=b^ab^ab^{-1}[a,b]^{n-1}b^{-1}= (b^2)^ab^{-2}b[a,b]^{n-1}b^{-1}=[a,b^2](b[a,b]b^{-1})^{n-1}=[a,b^2]([a,b]^b)^{n-1}

=[a,b^2][a^b,b^b]^{n-1}=[a,b^2][a^b,b]^{n-1}. \ \Box

Problem 2. Let C=\{[a,b]: \ a,b \in G\}. If [G:Z(G)]=n, then |C| \leq n^2.

Proof. Define the map \varphi : C \longrightarrow G/Z(G) \times G/Z(G) by \varphi([a,b])=(aZ(G),bZ(G)). If we prove that \varphi is one-to-one, we are done because then |C| \leq |G/Z(G) \times G/Z(G)|=n^2. So suppose that \varphi([a,b])=\varphi([c,d]). Then aZ(G)=cZ(G) and bZ(G)=dZ(G) and hence a^{-1}c \in Z(G) and b^{-1}d \in Z(G). Therefore

[a,b]=aba^{-1}b^{-1}=ab(a^{-1}c)c^{-1}b^{-1} =a(a^{-1}c)bc^{-1}b^{-1}=cbc^{-1}b^{-1} =cbc^{-1}(b^{-1}d)d^{-1}

=cb(b^{-1}d)c^{-1}d^{-1}=cdc^{-1}d^{-1}=[c,d]. \ \Box

Schur’s theorem. If [G:Z(G)]=n, then \displaystyle |G'| \leq n^{2n^3}.

Solution. Let C=\{[a,b]: \ a,b \in G\} and c \in G'. Then c = c_1c_2 \ldots c_m, where c_i \in C. We will choose the integer m to be as small as possible, i.e. if c = c_1'c_2' \ldots c_k', with c_i' \in C, then k \geq m. Now, we know from Problem 2, that the number of elements of C is at most n^2. So if we prove that each c_i can occur at most n times in c = c_1c_2 \cdots c_m, then m \leq n|C| \leq n^3 and thus there will be at most (n^2)^m \leq n^{2n^3} possible values for c and the problem is solved. To prove that each c_i can occur at most n times in c = c_1c_2 \ldots c_m, suppose, to the contrary, that, say c_j, occurs r \geq n+1 times in the product. Then by part 2) of Problem 1, we can move each c_j to the right hand side of the product to get c = c_1'c_2' \ldots c_s' c_j^r, where c_i' \in C and r + s = m. But by part 3) of Problem 1, c_j^{n+1} is a product of n elements of C and hence, since c_j^r=c_j^{r-(n+1)}c_j^{n+1}, we see that c_j^r is a product of r-1 elements of C. Therefore c is a product of s+r-1=m-1 elements of C, which contradicts the minimality of m. \ \Box

If G is finitely generated, then the converse of Schur’s theorem is also true, i.e. if G' is finite, then G/Z(G) is finite too. It’s not hard, try to prove it!

Here you can see part (2). We are now ready to prove a nice result.

Theorem. Let k be an algebraically closed field. Let A be a commutative k-domain and let B be a k-domain. Then A \otimes_k B is a k-domain.

Proof. Suppose that A \otimes_k B is not a domain. So there exist non-zero elements u, v \in A \otimes_k B such that uv=0. Let u = \sum_{i=1}^m a_i \otimes_k b_i and v = \sum_{i=1}^n a_i' \otimes_k b_i', where a_i,a_i' \in A, \ b_i,b_i' \in B and both sets \{b_1, \ldots , b_m\} and \{b_1', \ldots , b_n'\} are k-linearly independent. Let C=k[a_1, \ldots , a_m, a_1', \ldots , a_n'], which is a commutative domain because C \subseteq A. Also, note that u,v \in C \otimes_k B \subseteq A \otimes_k B. Now,  since u, v \neq0, there exist integers r,s such that a_r \neq 0 and a_s' \neq 0. Therefore, by the third part of the corollary in part (2), there exists a ring homomorphism \varphi : C \longrightarrow k such that \varphi(a_r) \neq 0 and \varphi(a_s') \neq 0. Let \psi = \varphi \otimes \text{id}_B. Then \psi : C \otimes_k B \longrightarrow k \otimes_k B \cong B is a ring homomorphism and hence \psi(u)\psi(v)=\psi(uv)=0. Therefore either \psi(u)=0 or \psi(v)=0, because B is a domain. But \psi(u)=\sum_{i=1}^m \varphi(a_i) \otimes_k b_i \neq 0, because \{b_1, \ldots , b_m \} is k-linearly independent and \varphi(a_r) \neq 0. Also, \psi(v)=\sum_{i=1}^n \varphi(a_i') \otimes_k b_i' \neq 0, because \{b_1', \ldots , b_n' \} is k-linearly independent and \varphi(a_s') \neq 0. This contradiction proves that A \otimes_k B is  a domain. \Box

Let k be an algebraically closed field. A trivial corollary of the theorem is a well-known result in field theory: if F_1,F_2 are two fields which contain k, then F_1 \otimes_k F_2 is a commutative domain. Another trivial result is this: if k is contained in both a field F and the center of a division algebra D, then F \otimes_k D is a domain.

Question. Let k be an algebraically closed field and let D_1,D_2 be two finite dimensional division k-algebras. Will D_1 \otimes_k D_2 always be a domain?

Answer. No! See the recent paper of Louis Rowen and David Saltman for an example.