Tensor Products of Z-Modules II (Tensor Product of Two Fields over the Integers)

Case 1: Fields with Different Characteristics.

Theorem (Different Characteristics). Let $L$ and $K$ be fields. If $\mathrm{char}(L)\ne \mathrm{char}(K)$, then $L{\otimes }_{\mathbb{Z}}K\cong \{0\}$.

Proof. Let $\mathrm{char}(L)=p>0$ and $\mathrm{char}(K)=q$, where $q=0$ or $q$ is a prime distinct from $p$. For any simple tensor $l\otimes k\in L{\otimes }_{\mathbb{Z}}K$, we have $p\cdot (l\otimes k)=(pl)\otimes k=0\otimes k=0$.

If $q=0$, then $K$ is a field of characteristic zero, so $p$ is an invertible element in $K$. Any $k\in K$ can be written as $k=p\cdot (k/p)$.

$$l\otimes k=l\otimes (p\cdot (k/p))=(p\cdot l)\otimes (k/p)=0\otimes (k/p)=0$$

If $q>0$ and $q\ne p$, then $p$ and $q$ are distinct primes, so $\gcd (p,q)=1$. By the Bézout identity, there exist integers $a,b$ such that $ap+bq=1$. For any simple tensor $x=l\otimes k$, we have $p\cdot x=(pl)\otimes k=0$ and $q\cdot x=l\otimes (qk)=0$.

$$x=1\cdot x=(ap+bq)\cdot x=a(p\cdot x)+b(q\cdot x)=a\cdot 0+b\cdot 0=0$$

Since every simple tensor is zero and these elements generate the module, $L{\otimes }_{\mathbb{Z}}K\cong \{0\}$. $\square$

Case 2: Fields with the Same Characteristic.

Theorem (Same Characteristic). Let $L$ and $K$ be fields with the same characteristic. Let $F$ be their common prime subfield ($F=\mathbb{Q}$ if characteristic is 0, $F={\mathbb{F}}_p$ if characteristic is $p$). There is a ring isomorphism:

$$L{\otimes }_{\mathbb{Z}}K\cong L{\otimes }_{F}K$$

Proof.

The fields $L$ and $K$ are $F$-algebras, so the tensor product $L{\otimes }_{F}K$ is a well-defined $F$-algebra.

Consider the canonical map $\varphi :L\times K\to L{\otimes }_{F}K$ given by $\varphi (l,k)=l\otimes k$. This map is $F$-bilinear by definition, and therefore also $\mathbb{Z}$-bilinear since the action of any $n\in \mathbb{Z}$ is a special case of the action of $F$. By the universal property of the tensor product over $\mathbb{Z}$, $\varphi$ induces a unique $\mathbb{Z}$-module homomorphism $\Phi :L{\otimes }_{\mathbb{Z}}K\to L{\otimes }_{F}K$ such that $\Phi (l\otimes k)=l\otimes k$ for all $l\in L,k\in K$.

Conversely, consider the map $\psi :L\times K\to L{\otimes }_{\mathbb{Z}}K$ given by $\psi (l,k)=l\otimes k$. This map is $\mathbb{Z}$-bilinear. We show it is also $F$-bilinear. Let $\alpha \in F$. By symmetry, it suffices to show that for any $l\in L,k\in K$, the relation $(\alpha \cdot l)\otimes k=l\otimes (\alpha \cdot k)$ holds in $L{\otimes }_{\mathbb{Z}}K$.

Case 1: The characteristic is $p>0$. Then $F={\mathbb{F}}_p$, and any $\alpha \in F$ can be represented by an integer $n$ where $0\le n<p$. The action is $\alpha \cdot l=n\cdot l$. Then $(\alpha \cdot l)\otimes k=(n\cdot l)\otimes k=n(l\otimes k)=l\otimes (n\cdot k)=l\otimes (\alpha \cdot k)$.

Case 2: The characteristic is 0. Then $F=\mathbb{Q}$, and any $\alpha \in F$ can be written as $n/m$ for $n,m\in \mathbb{Z}$ with $m\ne 0$. For any integer $n$, $(n\cdot l)\otimes k=n(l\otimes k)=l\otimes (n\cdot k)$. For the reciprocal $1/m$, we have $m\cdot ((1/m\cdot l)\otimes k)=(m\cdot (1/m\cdot l))\otimes k=l\otimes k$. Similarly, $m\cdot (l\otimes (1/m\cdot k))=l\otimes (m\cdot (1/m\cdot k))=l\otimes k$. Since $L$ and $K$ are $\mathbb{Q}$-vector spaces, they are divisible $\mathbb{Z}$-modules, which makes $L{\otimes }_{\mathbb{Z}}K$ a divisible $\mathbb{Z}$-module. Thus, multiplication by a non-zero integer $m$ is an automorphism. It follows that $(1/m\cdot l)\otimes k=l\otimes (1/m\cdot k)$. Combining these results gives $((n/m)\cdot l)\otimes k=n((1/m\cdot l)\otimes k)=n(l\otimes (1/m\cdot k))=l\otimes (n(1/m\cdot k))=l\otimes ((n/m)\cdot k)$.

In both cases, $\psi$ is $F$-bilinear. By the universal property of the tensor product over $F$, $\psi$ induces a unique $F$-module homomorphism $\Psi :L{\otimes }_{F}K\to L{\otimes }_{\mathbb{Z}}K$ such that $\Psi (l\otimes k)=l\otimes k$.

The maps $\Phi$ and $\Psi$ are mutually inverse, since $(\Psi \circ \Phi )(l\otimes k)=\Psi (l\otimes k)=l\otimes k$ and $(\Phi \circ \Psi )(l\otimes k)=\Phi (l\otimes k)=l\otimes k$. As these compositions are the identity on the generating simple tensors, they are the identity maps on the respective modules. Thus, $\Phi$ is an isomorphism of abelian groups.

Finally, $\Phi$ is a ring homomorphism:

$$\Phi ((l_1\otimes k_1)(l_2\otimes k_2))=\Phi ((l_1{l}_2)\otimes (k_1{k}_2))=(l_1{l}_2)\otimes (k_1{k}_2)=(l_1\otimes k_1)(l_2\otimes k_2)=\Phi (l_1\otimes k_1)\Phi (l_2\otimes k_2)$$

Therefore, $\Phi$ is a ring isomorphism. $\square$

The structure of $L{\otimes }_{F}K$ is determined by the behavior of one field extension over the other. Assume $L/F$ is a finite extension. By the Primitive Element Theorem, $L\cong F[x]/(f(x))$ for some irreducible polynomial $f(x)\in F[x]$. This gives the isomorphism:

$$L{\otimes }_{F}K\cong (F[x]/(f(x))){\otimes }_{F}K\cong K[x]/(f(x))$$

Let the prime factorization of $f(x)$ in the polynomial ring $K[x]$ be $f(x)=g_1(x{)}^{e_1}\cdots g_r(x{)}^{e_r}$. By the Chinese Remainder Theorem:

$$K[x]/(f(x))\cong K[x]/(g_1(x{)}^{e_1})\times \cdots \times K[x]/(g_r(x{)}^{e_r})$$

Subcase: Separable Extensions.

If $L/F$ is a separable extension, then $f(x)$ is a separable polynomial, which implies all exponents in its factorization over $K$ are $e_i=1$. Each factor $K[x]/(g_i(x))$ is a field, as each $g_i(x)$ is irreducible over $K$. Thus, for separable extensions, $L{\otimes }_{F}K$ is a direct product of fields. It is a field if and only if $f(x)$ remains irreducible over $K$ (i.e., $r=1$).

Example (Tensor product is a field). Consider $\mathbb{Q}(\sqrt{2}){\otimes }_{\mathbb{Z}}\mathbb{Q}(\sqrt{3})$. The common prime subfield is $F=\mathbb{Q}$. Let $L=\mathbb{Q}(\sqrt{2})\cong \mathbb{Q}[x]/(x^2-2)$ and $K=\mathbb{Q}(\sqrt{3})$. The tensor product is isomorphic to $\mathbb{Q}(\sqrt{3})[x]/(x^2-2)$. The polynomial $x^2-2$ is irreducible over $\mathbb{Q}(\sqrt{3})$, so the resulting ring is a field, which is the compositum $\mathbb{Q}(\sqrt{2},\sqrt{3})$.

Example (Tensor product is a product of fields). Consider $\mathbb{C}{\otimes }_{\mathbb{Z}}\mathbb{C}$. This is isomorphic to $\mathbb{C}{\otimes }_{\mathbb{Q}}\mathbb{C}$. It is easier to view this as $\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{C}$ since both are $\mathbb{R}$-algebras. Let $L=\mathbb{C}\cong \mathbb{R}[x]/(x^2+1)$ and $K=\mathbb{C}$. The tensor product is isomorphic to $\mathbb{C}[x]/(x^2+1)$. Over $\mathbb{C}$, the polynomial factors as $x^2+1=(x-i)(x+i)$. By the Chinese Remainder Theorem:

$$\mathbb{C}[x]/(x^2+1)\cong \mathbb{C}[x]/(x-i)\times \mathbb{C}[x]/(x+i)\cong \mathbb{C}\times \mathbb{C}$$

This ring is not a field as it contains zero divisors, e.g., $(1,0)\cdot (0,1)=(0,0)$.

Subcase: Inseparable Extensions.

This case occurs only when $\mathrm{char}(F)=p>0$. If $L/F$ is an inseparable extension, the polynomial $f(x)$ is not separable, and at least one exponent $e_i$ in its factorization over $K$ will be greater than 1. The ring $K[x]/(g_i(x{)}^{e_i})$ is not a field and contains non-zero nilpotent elements. For instance, the element represented by the class of $g_i(x)$ is non-zero, but its $e_i$-th power is zero.

Example (Tensor product with nilpotents). Let $F={\mathbb{F}}_p(t)$ be the field of rational functions in $t$ over ${\mathbb{F}}_p$. Let $L=F(t^{1/p})$. This is an inseparable extension, as $L\cong F[x]/(x^p-t)$. Consider the tensor product $L{\otimes }_{F}L$.

$$L{\otimes }_{F}L\cong L[x]/(x^p-t)$$

In the field $L$, the element $t$ has a $p$-th root, namely $t^{1/p}$. The polynomial factors completely over $L$:

$$x^p-t=x^p-(t^{1/p}{)}^p=(x-t^{1/p}{)}^p$$

Therefore, $L{\otimes }_{F}L\cong L[x]/((x-t^{1/p}{)}^p)$. Let $y=x-t^{1/p}$. The ring is isomorphic to $L[y]/(y^p)$. In this ring, the element represented by $y$ is non-zero, but $y^p=0$, so it is a non-zero nilpotent element.