## Tensor product of division algebras (1)

Posted: September 1, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , ,

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.

## Tensor product does not distribute over direct product

Posted: February 21, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,

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). $\Box$

## Direct limit of modules & tensor product

Posted: February 16, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
Tags: ,

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 1$R$ 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).

## Tensor product of simple algebras need not be simple

Posted: January 29, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: ,

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 $xy=0. \ \Box$

## Tensor product of simple algebras (4)

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: ,

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 2$S \cong M_n(\mathbb{C}),$ 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 $n$th 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.