Tensor Products

We need only a special case of the tensor product construction. Let $A$ and $B$ be commutative rings containing a field $K$ and suppose that $B$ is of finite dimension $N$ over $K$, say, with basis

$$1=w_1, w_2, \ldots, w_N.$$

Then $B$ is determined up to isomorphism as a ring over $K$ by the multiplication table $(c_{i,j,n})$ defined by

$$w_i \cdot w_j = \sum_{n=1}^N c_{i,j,n} \cdot w_n.$$

We define a new ring $C$ containing $K$ whose elements are the set of all expressions

$$\sum_{n=1}^N a_n \underline{w}_n$$

where the $\underline{w}_n$ have the same multiplication rule

$$\underline{w}_i \cdot \underline{w}_j = \sum_{n=1}^N c_{i,j,n} \cdot \underline{w}_n$$

as the $w_n$.

There are injective ring homomorphisms

$$i:A\hookrightarrow C, \qquad i(a) = a \underline{w}_1$$   (note that $w_1=1$)

and

$$j:B\hookrightarrow C, \qquad j\left(\sum_{n=1}^N c_n w_n\right) = \sum_{n=1}^N c_n \underline{w}_n. \qquad\quad$$  

Moreover $C$ is defined, up to isomorphism, by $A$ and $B$ and is independent of the particular choice of basis $w_n$ of $B$ (i.e., a change of basis of $B$ induces a canonical isomorphism of the $C$ defined by the first basis to the $C$ defined by the second basis). We write

$$C = A\otimes _K B$$

since $C$ is, in fact, a special case of the ring tensor product.

Let us now suppose, further, that $A$ is a topological ring, i.e., has a topology with respect to which addition and multiplication are continuous. Then the map

$$C\to A \oplus \cdots \oplus A,\qquad \sum_{m=1}^N a_m \underline{w}_m \mapsto (a_1,\ldots, a_N)$$

defines a bijection between $C$ and the product of $N$ copies of $A$ (considered as sets). We give $C$ the product topology. It is readily verified that this topology is independent of the choice of basis $w_1, \ldots, w_N$ and that multiplication and addition on $C$ are continuous, so $C$ is a topological ring. We call this topology on $C$ the tensor product topology.

Now drop our assumption that $A$ and $B$ have a topology, but suppose that $A$ and $B$ are not merely rings but fields. Recall that a finite extension $L/K$ of fields is separable if the number of embeddings $L\hookrightarrow \overline{K}$ that fix $K$ equals the degree of $L$ over $K$, where $\overline{K}$ is an algebraic closure of $K$. The primitive element theorem from Galois theory asserts that any such extension is generated by a single element, i.e., $L=K(a)$ for some $a\in L$.

Lemma 16.2.1   Let $A$ and $B$ be fields containing the field $K$ and suppose that $B$ is a separable extension of finite degree $N=[B:K]$. Then $C=A\otimes _K B$ is the direct sum of a finite number of fields $K_j$, each containing an isomorphic image of $A$ and an isomorphic image of $B$.

Proof. By the primitive element theorem, we have $B=K(b)$, where $b$ is a root of some separable irreducible polynomial $f(x)\in K[x]$ of degree $N$. Then $1,b,\ldots, b^{N-1}$ is a basis for $B$ over $K$, so

$$A\otimes _K B = A[\underline{b}] \cong A[x]/(f(x))$$

where $1,\underline{b},\underline{b}^2,\ldots,\underline{b}^{N-1}$ are linearly independent over $A$ and $\underline{b}$ satisfies $f(\underline{b})=0$.

Although the polynomial $f(x)$ is irreducible as an element of $K[x]$, it need not be irreducible in $A[x]$. Since $A$ is a field, we have a factorization

$$f(x) = \prod_{j=1}^J g_j(x)$$

where $g_j(x)\in A[x]$ is irreducible. The $g_j(x)$ are distinct because $f(x)$ is separable (i.e., has distinct roots in any algebraic closure).

For each $j$, let $\underline{b}_j\in \overline{A}$ be a root of $g_j(x)$, where $\overline{A}$ is a fixed algebraic closure of the field $A$. Let $K_j = A(\underline{b}_j)$. Then the map

$$\varphi _j : A\otimes _K B \to K_j$$ (16.1)

given by sending any polynomial $h(\underline{b})$ in $\underline{b}$ (where $h\in A[x]$) to $h(\underline{b}_j)$ is a ring homomorphism, because the image of  $\underline{b}$ satisfies the polynomial $f(x)$, and $A\otimes _K B\cong A[x]/(f(x))$.

By the Chinese Remainder Theorem, the maps from (16.2.1) combine to define a ring isomorphism

$$A\otimes _K B \cong A[x]/(f(x)) \cong \bigoplus_{j=1}^J A[x]/(g_j(x)) \cong \bigoplus_{j=1}^J K_j.$$

Each $K_j$ is of the form $A[x]/(g_j(x))$, so contains an isomorphic image of $A$. It thus remains to show that the ring homomorphisms

$$\lambda_j : B \xrightarrow{b \mapsto 1\otimes b} A\otimes _K B \xrightarrow{\varphi _j} K_j$$

are injections. Since $B$ and $K_j$ are both fields, $\lambda_j$ is either the 0 map or injective. However, $\lambda_j$ is not the 0 map since $\lambda_j(1)=1\in K_j$. $\qedsymbol$

Example 16.2.2   If $A$ and $B$ are finite extensions of $\mathbf {Q}$, then $A\otimes _\mathbf{Q}B$ is an algebra of degree $[A:\mathbf{Q}]\cdot [B:\mathbf{Q}]$. For example, suppose $A$ is generated by a root of $x^2+1$ and $B$ is generated by a root of $x^3-2$. We can view $A\otimes _\mathbf{Q}B$ as either $A[x]/(x^3-2)$ or $B[x]/(x^2+1)$. The polynomial $x^2+1$ is irreducible over $\mathbf {Q}$, and if it factored over the cubic field $B$, then there would be a root of $x^2+1$ in $B$, i.e., the quadratic field $A=\mathbf{Q}(i)$ would be a subfield of the cubic field $B=\mathbf{Q}(\sqrt[3]{2})$, which is impossible. Thus $x^2+1$ is irreducible over $B$, so $A\otimes _\mathbf{Q}B = A.B = \mathbf{Q}(i,\sqrt[3]{2})$ is a degree $6$ extension of $\mathbf {Q}$. Notice that $A.B$ contains a copy $A$ and a copy of $B$. By the primitive element theorem the composite field $A.B$ can be generated by the root of a single polynomial. For example, the minimal polynomial of $i+\sqrt[3]{2}$ is $x^6 + 3x^4 - 4x^3 + 3x^2 + 12x + 5$, hence $\mathbf{Q}(i+\sqrt[3]{2})=A.B$.

Example 16.2.3   The case $A\cong B$ is even more exciting. For example, suppose $A=B=\mathbf{Q}(i)$. Using the Chinese Remainder Theorem we have that

$$\mathbf{Q}(i)\otimes _\mathbf{Q}\mathbf{Q}(i) \cong \mathbf{Q}(i)... ... \cong \mathbf{Q}(i)[x]/((x-i)(x+i)) \cong \mathbf{Q}(i) \oplus \mathbf{Q}(i),$$

since $(x-i)$ and $(x+i)$ are coprime. The last isomorphism sends $a + b x$, with $a,b\in\mathbf{Q}(i)$, to $(a+bi, a-bi)$. Since $\mathbf{Q}(i)\oplus \mathbf{Q}(i)$ has zero divisors, the tensor product $\mathbf{Q}(i)\otimes _\mathbf{Q}\mathbf{Q}(i)$ must also have zero divisors. For example, $(1,0)$ and $(0,1)$ is a zero divisor pair on the right hand side, and we can trace back to the elements of the tensor product that they define. First, by solving the system

$$a+bi=1$$    and $$\qquad a-bi=0$$

we see that $(1,0)$ corresponds to $a=1/2$ and $b=-i/2$, i.e., to the element

$$\frac{1}{2}- \frac{i}{2} x\in \mathbf{Q}(i)[x]/(x^2+1).$$

This element in turn corresponds to

$$\frac{1}{2}\otimes 1 - \frac{i}{2}\otimes i \in \mathbf{Q}(i)\otimes _\mathbf{Q}\mathbf{Q}(i).$$

Similarly the other element $(0,1)$ corresponds to

$$\frac{1}{2}\otimes 1 + \frac{i}{2}\otimes i \in \mathbf{Q}(i)\otimes _\mathbf{Q}\mathbf{Q}(i).$$

As a double check, observe that

$$\left(\frac{1}{2}\otimes 1 - \frac{i}{2}\otimes i\right)\cdot \left(\frac{1}{2}\otimes 1 + \frac{i}{2}\otimes i\right) = \frac{1}{4}\otimes 1 + \frac{i}{4}\otimes i - \frac{i}{4}\otimes i -\frac{i^2}{4}\otimes i^2$$

$$= \frac{1}{4}\otimes 1 - \frac{1}{4}\otimes 1 = 0 \in \mathbf{Q}(i)\otimes _\mathbf{Q}\mathbf{Q}(i).$$

Clearing the denominator of $2$ and writing $1\otimes 1 = 1$, we have $(1-i\otimes i)(1+i\otimes i) = 0$, so $i\otimes i$ is a root of the polynomimal $x^2-1$, and $i\otimes i$ is not $\pm 1$, so $x^2-1$ has more than $2$ roots.

In general, to understand $A\otimes _K B$ explicitly is the same as factoring either the defining polynomial of $B$ over the field $A$, or factoring the defining polynomial of $A$ over $B$.

Corollary 16.2.4   Let $a\in B$ be any element and let $f(x)\in K[x]$ be the characteristic polynomials of $a$ over $K$ and let $g_j(x)\in A[x]$ (for $1\leq j \leq J$) be the characteristic polynomials of the images of $a$ under $B\to A\otimes _K B \to K_j$ over $A$, respectively. Then

$$f(x) = \prod_{j=1}^J g_j(X).$$ (16.2)

Proof. We show that both sides of (16.2.2) are the characteristic polynomial $T(x)$ of the image of $a$ in $A\otimes _K B$ over $A$. That $f(x)=T(x)$ follows at once by computing the characteristic polynomial in terms of a basis $\underline{w}_1,\ldots, \underline{w}_N$ of $A\otimes _K B$, where $w_1, \ldots, w_N$ is a basis for $B$ over $K$ (this is because the matrix of left multiplication by $b$ on $A\otimes _K B$ is exactly the same as the matrix of left multiplication on $B$, so the characteristic polynomial doesn't change). To see that $T(X) = \prod g_j(X)$, compute the action of the image of $a$ in $A\otimes _K B$ with respect to a basis of

$$A\otimes _K B \cong \bigoplus_{j=1}^J K_j$$ (16.3)

composed of basis of the individual extensions $K_j$ of $A$. The resulting matrix will be a block direct sum of submatrices, each of whose characteristic polynomials is one of the $g_j(X)$. Taking the product gives the claimed identity (16.2.2). $\qedsymbol$

Corollary 16.2.5   For $a\in B$ we have

$$\Norm _{B/K}(a) = \prod_{j=1}^J \Norm _{K_j/A}(a),$$

and

$$\Tr _{B/K}(a) = \sum_{j=1}^J \Tr _{K_j/A}(a),$$

Proof. This follows from Corollary 16.2.4. First, the norm is $\pm$ the constant term of the characteristic polynomial, and the constant term of the product of polynomials is the product of the constant terms (and one sees that the sign matches up correctly). Second, the trace is minus the second coefficient of the characteristic polynomial, and second coefficients add when one multiplies polynomials:

$$(x^n + a_{n-1}x^{n-1} + \cdots ) \cdot (x^m + a_{m-1}x^{m-1} + \cdots ) = x^{n+m} + x^{n+m-1} (a_{m-1} + a_{n-1}) + \cdots.$$

One could also see both the statements by considering a matrix of left multiplication by $a$ first with respect to the basis of $\underline{w}_n$ and second with respect to the basis coming from the left side of (16.2.3).

$\qedsymbol$