Posts Tagged ‘tensor product’

In this post we showed that \mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \cong M_4(\mathbb{R}). So the tensor product of two (finite dimensional) division algebras may not even be a reduced algebra let alone a division algebra. There is however the following simple result.

Theorem. Let D_1, D_2 be finite dimensional central division k-algebras. If \gcd(\dim_k D_1, \dim_k D_2)=1, then D_1 \otimes_k D_2 is a division algebra.

Proof. By the corollary in this post, D_1 \otimes_k D_2 is a finite dimensional central simple k-algebra. So

D_1 \otimes_k D_2 \cong M_n(D) \ \ \ \ \ \ \ \ \ \ (1)

for some finite dimensional central division k-algebra D and some integer n \geq 1.  We need to prove that n=1. Let \dim_k D_i=m_i, \ i=1,2. We are given that

\gcd(m_1,m_2)=1. \ \ \ \ \ \ \ \ \ \ (2)

Now, we have

D_i^{op} \otimes_k D \cong M_{r_i}(D_i'), \ \ i=1,2, \ \ \ \ \ \ \ \ \ (3)

for some finite dimensional central division k-algebras D_i' and integers r_i \geq 1. Also, by the theorem in this post,

D_i^{op} \otimes_k D_i \cong M_{m_i}(k), \ i=1,2. \ \ \ \ \ \ \ \ \ \ (4)

Thus, using (1), (3) and (4), we have

M_{m_1}(D_2) \cong M_{m_1}(k) \otimes_k D_2 \cong D_1^{op} \otimes_k D_1 \otimes_k D_2 \cong D_1^{op} \otimes_k M_n(D) \cong

M_n(D_1^{op} \otimes_k D) \cong M_n(M_{r_1}(D_1')) \cong M_{nr_1}(D_1').

Hence m_1=nr_1. Similarly M_{m_2}(D_1) \cong M_{nr_2}(D_2') and so m_2=nr_2. So n divides both m_1 and m_2 and hence n=1, by (2). \ \Box

See part (2) here.


Problem 1. Let G be an abelian group. Prove that if G has no element of infinite order,  then \mathbb{Q} \otimes_{\mathbb{Z}}G=(0).

Solution.  If g \in G and o(g)=n, then for all r \in \mathbb{Q} we have

r \otimes_{\mathbb{Z}}g=(nr/n) \otimes_{\mathbb{Z}}g=(r/n) \otimes_{\mathbb{Z}}(ng)=(r/n) \otimes_{\mathbb{Z}}0=0. \Box

Problem 2. Let G be an abelian group. Prove that if G has an element of infinite order, then \mathbb{Q} \otimes_{\mathbb{Z}} G \neq (0).

Solution. Let g be an element of infinite order in G. Since \mathbb{Q} is a flat \mathbb{Z}-module, \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle is a subgroup of \mathbb{Q} \otimes_{\mathbb{Z}} G. So we only need to prove that \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle \neq (0). This is obvious because, since g has infinite order, we have \langle g \rangle \cong \mathbb{Z} and thus \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle \cong \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z} \cong \mathbb{Q} \neq (0). \ \Box

Let R be a commutative ring with unity and let I be a set which may or may not be finite. Suppose that  Y_i, \ i\in I, are R-modules. Let X be an R-module. Recall that tensor product distributes over direct sum, i.e.

X \otimes_R \bigoplus_{i \in I} Y_i \cong \bigoplus_{i \in I}(X \otimes_R Y_i),

as R-modules. Next problem shows that the above is not true in general if \bigoplus is replaced with \prod.

Problem 3. Let G_i = \mathbb{Z}/2^i\mathbb{Z}, \ i\in \mathbb{N}. Prove that \mathbb{Q} \otimes_{\mathbb{Z}} \prod_{i=1}^{\infty} G_i \ncong \prod_{i=1}^{\infty} (\mathbb{Q} \otimes_{\mathbb{Z}}G_i).

Solution. Each G_i is a finite abelian group and thus \mathbb{Q} \otimes_{\mathbb{Z}} G_i = (0), by Problem 1. Thus

\prod_{i=1}^{\infty} (\mathbb{Q} \otimes_{\mathbb{Z}} G_i)=(0). \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

Now let G = \prod_{i=1}^{\infty}G_i. Let g = (1 + 2^i \mathbb{Z})_{i \in \mathbb{N}} \in G. Suppose that the order of g is finite, say n. Then n + 2^i \mathbb{Z} = 0, for all i \in \mathbb{N}. That means  2^i divides n for all i \in \mathbb{N}, which is nonsense. So the order of g in G is infinite and hence, by Problem 2,

\mathbb{Q} \otimes_{\mathbb{Z}} G \neq (0). \ \ \ \ \ \ \ \ \ \ \ \ (2)

The result now follows from (1) and (2). \Box

Theorem. Let R be a commutative ring, A an R-module and \{M_i, f_{ij} \} a direct system of R-modules. Then

\varinjlim (A \otimes_R M_i) \cong A \otimes_R \varinjlim M_i.

The above isomorphism is an isomorphism of R-modules. We will give the proof of this theorem at the end of this post. So let’s get prepared for the proof!

Notation 1R is a commutative ring, A is an R-module and \{M_i, f_{ij} \} is a direct system of R-modules over some partially ordered index set I. We will not assume that I is directed.

Notation 2. We proved in here that M=\varinjlim M_i exists. Recall that, as a part of the definition of direct limit, we also have canonical homomorphisms f_i : M_i \longrightarrow M satisfying the relations f_jf_{ij}=f_i, for all i \leq j. For modules we explicity defined f_i by f_i(x_i)=\rho_i(x_i) + N (see the theorem in here), but we will not be needing that.

Notation 3. Let A be an R-module. For all i,j \in I with i \leq j we will let M_i' = A \otimes_R M_i and f'_{ij}=\text{id}_A \otimes_R f_{ij}. By Example 5, \{M_i', f'_{ij} \} is a direct system of R-modules and so M'=\varinjlim (A \otimes_R M_i) exists. We also have canonical homomorphisms f_i' : M_i' \longrightarrow M' satisfying f_j'f_{ij}'=f_i', for all i \leq j.

Notation 4. For every i \in I we will let h_i = \text{id}_A \otimes f_i.

Remark 1.  Clearly h_i : M'_i \longrightarrow A \otimes_R M and h_jf_{ij}'=h_i, for all i \leq j, because

h_j f_{ij}' = (\text{id}_A \otimes f_j)(\text{id}_A \otimes f_{ij})=\text{id}_A \otimes f_jf_{ij}=\text{id}_A \otimes f_i = h_i.

 Lemma. Let X be an R-module. Suppose that for every i \in I there exists an R-module homomorphism g_i : M_i' \longrightarrow X such that g_jf_{ij}' = g_i, for all i \leq j. Let a \in A. Then

1) There exist R-module homomorphisms \nu_{i,a} : M_i \longrightarrow X such that \nu_{j,a} f_{ij}=\nu_{i,a}, for all i \leq j.

2) There exists a unique R-module homomorphism \nu_a : M \longrightarrow X such that \nu_a f_i = v_{i,a}, for all i \in I.

Proof. 1) \nu_{i,a} are defined very naturally: \nu_{i,a}(x_i)=g_i(a \otimes_R x_i), for all x_i \in M_i. Then

\nu_{j,a} f_{ij}(x_i)=g_j(a \otimes_R f_{ij}(x_i))=g_j f_{ij}'(a \otimes_R x_i)=g_i(a \otimes_R x_i)=\nu_{i,a}(x_i).

2) This part is obvious from the first part and the universal property of direct limit. (See Definition 1 in here) \Box

Proof of the Theorem. We will show that A \otimes_R M satisfies the conditions in the definition of M' = \varinjlim M_i' (see Definition 1 in here) and thus, by the uniqueness of direct limit, A \otimes_R M \cong M' and the theorem is proved. The first condition is satisfied by Remark 1. For the second condition (the universal property), suppose that X is an R-module and g_i : M_i' \longrightarrow X are R-module homomorphisms such that g_jf_{ij}'=g_i, whenever i \leq j. So the hypothesis in the above lemma is satisfied. For a \in A and i \in I let \nu_{i,a} and \nu_a be the maps in the lemma. Define the map \nu : A \times M \longrightarrow X by v(a,x)=v_a(x), for all a \in A and x \in M. See that \nu is R-bilinear and so it induces an R-module homomorphism f : A \otimes_R M \longrightarrow X defined by f(a \otimes_R x)=v_a(x). We also have

 fh_i(a \otimes_R x_i)=f(a \otimes_R f_i(x_i))=\nu_af_i(x_i)=\nu_{i,a}(x_i)=g_i(a \otimes_R x_i),

for all a \in A, \ i \in I and x_i \in M_i. Thus fh_i = g_i. So the only thing left is the uniqueness of f, which is obvious because, as we mentioned in the second part of the above lemma, \nu_a is uniquely defined for a given a \in A. \ \Box

Remark 2. If R is not commutative, M_i are left R-modules and A is a right R-module (resp. (R,R)-bimodule), then the isomorphism in the theorem is an isomorphism of abelian groups (resp. left R-modules).

By Corollary 1, if A,B are simple k-algebras and the center of A or B is k, then A \otimes_k B is simple. The condition that k is the center of A or B cannot be omited, as the following examples show.

Example 1. Let p be a prime number, F :=\mathbb{Z}/p\mathbb{Z} and R:=F(x) \otimes_{F(x^p)} F(x), where  F(x) is the field of rational functions in x over F. Then R is not simple.

Proof. Since a commutative simple ring is a field, we just need to show that R is not a field. We will do that by proving that R has a non-zero nilpotent element and thus R is not even a domain! Let y=x \otimes 1 - 1 \otimes x. The set \{1,x \} is linearly independent over F(x^p) because otherwise we would have f(x^p)=xg(x^p), for  some f,g \in F[x], which is absurd. So y \neq 0, by Lemma 1. We now show that y^p=0, which will complete the proof: note that since x^p \in F(x^p), we have x^p \otimes u = 1 \otimes x^pu for all u \in F(x). Thus

y^p = x^p \otimes 1 - 1 \otimes x^p = 1 \otimes x^p - 1 \otimes x^p=0. \ \Box

Example 2. R := \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} is not simple.

Proof. Again, if R was simple, then it would be a field because it is commutative. We will show that R is not even a domain. Let x = 1 \otimes i - i \otimes 1 and y = 1 \otimes i + i \otimes 1. Then, both x,y are non-zero elements of R by Lemma 1 and the fact that \{1,i\} is linearly independent over \mathbb{R}. It is quickly seen that xy=0. \ \Box

Theorem. (Azumaya – Nakayama, 1947) Let A be a central simple k-algebra and let B be an arbitrary k-algebra. If J is a two-sided ideal of A \otimes_k B, then J = A \otimes_k (J \cap B).

Proof. It is obvious for J = (0) and so we may assume that J \neq (0). Let

I = J \cap B.

Consider the natural k-algebra homomorphism f: A \otimes_k B \longrightarrow A \otimes_k (B/I).

Claim 1. f(J) \cap (B/I) = (0). To see this, let y = 1 \otimes_k (b + I) \in f(J) \cap (B/I). Let \{a_i \} be a k-basis of A which contains 1. Let x \in J be such that f(x)=y. So x = \sum_{i=1}^n a_i \otimes_k b_i, for some integer n and b_i \in B. Then

y=1 \otimes_k (b+I)=f(x)=\sum_{i=1}^n a_i \otimes_k (b_i + I).

Thus for some i we have a_i = 1, \ b_i + I = b+I and b_j + I = 0, for all j \neq i. But then x = 1 \otimes_k b_i \in J \cap B = I and hence y = 1 \otimes_k (b+I)=1 \otimes_k (b_i + I) = 0.

Claim 2. J \subseteq \ker f. This is easy to see: since f is clearly onto, f(J) is a two-sided ideal of A \otimes_k (B/I). Thus, by the theorem we proved in part (3), we must have either f(J) \cap (B/I) \neq (0), which is not true by cliam 1, or f(J)=(0).

Claim 3. J \subseteq A \otimes_k I. This will be a result of claim 2 if we prove \ker f = A \otimes_k I. But it is obvious that A \otimes_k I \subseteq \ker f and so we only need to show that \ker f \subseteq A \otimes_k I. Let x \in \ker f. Let \{a_i \} be a k-basis for A. Then x = \sum_{i=1}^n a_i \otimes_k b_i, for some positive integer n and some b_i \in B. Then

0 = f(x) = \sum_{i=1}^n a_i \otimes_k (b_i + I)

and thus, since a_i are linearly independent, b_i + I = 0 for all i. Hence b_i \in I for all i and so x \in A \otimes_k I.

Claim 4. A \otimes_k I \subseteq J. The reason is that if a \in A, \ b \in I, then a \otimes_k b = (a \otimes_k 1)(1 \otimes_k b) \in J because 1 \otimes_k b \in I \subseteq J and J is an ideal of A \otimes_k B. \Box

Corollary. 1) If A is a central simple k-algebra and B is a simple k-algebra, then A \otimes_k B is a simple k-algebra.

2) If both A and B are central simple k-algebras, then A \otimes_k B is a central simple k-algebra.

Proof. Part 1) is a trivial result of the above theorem. Part 2) follows from part 1) and Lemma 2. \Box

Example. Let \mathbb{H} be the division algebra of quaternions over \mathbb{R}. Then \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \cong M_2(\mathbb{C}).

Proof. Let S=\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}. The first part of the above corollary implies that S is simple. Moreover, by Lemma 2, Z(S)=\mathbb{C} and so S is a central simple \mathbb{C}-algebra. Hence, by Remark 2S \cong M_n(\mathbb{C}), for some integer n \geq 1. Since \mathbb{C} is a field extension of \mathbb{R}, we have \dim_{\mathbb{C}}S = \dim_{\mathbb{R}} \mathbb{H}=4 and thus 4=\dim_{\mathbb{C}} M_n(\mathbb{C})=n^2. Hence n=2. \Box

Throughout k is a field, \text{char}(k)=0 and A is a k-algebra.

Example 2. Suppose that \delta is a derivation of A which is not inner. If A is \delta-simple, then B=A[x;\delta] is simple. In partcular, if A is simple, then A[x; \delta] is simple too.

Proof. Suppose, to the contrary, that B is not simple. So B has some non-zero ideal I \neq B. Let n be the minimum degree of non-zero elements of I. Let J be the set of leading coefficients of elements of I. Clearly J is a left ideal of A because I is an ideal of B. To see that J is also a right ideal of A, let a \in J and b \in A. Then there exists

f=ax^n + \text{lower degree terms} \in I.

But, by Remark 5

fb = abx^n + \text{lower degree terms} \in I

and so ab \in J, i.e. J is also a right ideal. Next, we’ll show that J is a \delta-ideal of A:

if a_n \in J, then there exists some f(x)=\sum_{i=0}^n a_ix^i \in I. Clearly xf - fx \in I, because I is an ideal of B. Now

xf - fx=\sum_{i=0}^n xa_i x^i - \sum_{i=0}^n a_ix^{i+1}=\sum_{i=0}^n (a_ix +\delta(a_i))x^i - \sum_{i=0}^n a_i x^{i+1}

= \sum_{i=0}^n \delta(a_i)x^i.

So \delta(a_n) \in J, i.e. J is a non-zero \delta-ideal of A. Therefore J=A, because A is \delta-simple. So 1 \in J, i.e. there exists g(x)=x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in I. Finally let a \in A. Now, ga - ag, which is an element of I, is a polynomial of degree at most n-1 and the coefficient of x^{n-1} is b_{n-1}a - ab_{n-1} + n \delta(a), which has to be zero because of the minimality of n. Thus, since \text{char}(k)=0, we may let c=\frac{b_{n-1}}{n} to get \delta(a)=ca-ac. That means \delta is inner. Contradiction! \Box

Lemma. If A is simple and \delta = \frac{d}{dx}, then A[x] is \delta-simple.

Proof. Let I \neq \{0\} be a \delta-ideal of A[x]. Let f=\sum_{i=0}^n a_ix^i, \ a_n \neq 0, be an element of I of the least degree. Suppose n > 0. Then, since I is a \delta-ideal, we must have \frac{df}{dx}=\sum_{i=0}^{n-1}ia_ix^{i-1} \in I. Hence na_{n-1}=0, by the minimality of n, and thus a_n=0 because \text{char}(k)=0.  This contradiction shows that n=0 and so f=a_0 \in A \cap I. Hence Aa_0A \subseteq I because I is an ideal of A[x] and A \subset A[x]. But A is simple and so Aa_0A = A, i.e. 1 \in Aa_0A \subseteq I and thus I=A[x]. \Box

Definition 5. The algebra A[x][y, \frac{d}{dx}] is called the first Weyl algebra over A and is denoted by \mathcal{A}_1(A). Inductively, for n \geq 2, we define \mathcal{A}_n(A)=\mathcal{A}_1(\mathcal{A}_{n-1}(A)) and we call \mathcal{A}_n(A) the nth Weyl algebra over A.

Example 3. If A is simple, then \mathcal{A}_n(A) is simple for all n. In particular, \mathcal{A}_n(k) is simple.

Proof. By Remark 3 in part (1), \delta = \frac{d}{dx} is a non-inner derivation of A[x]. So, \mathcal{A}_1(A) is simple by the above lemma and Example 2 in part (1). The proof is now completed by induction over n. \Box

Example 4. In this example we do not need to assume that \text{char}(k)=0. Let A be a simple ring and k its center. Let B be a simple k-algebra but the center of B may or may not be k. The first part of the corollary in this post shows that A \otimes_k B is simple. This is another way of constructing new simple rings from old ones.