Reference for tensor products of fields

Question

Does anybody know a reference for basic properties of tensor products of (finite) algebraic extensions of fields?

Ideally, I would like a description of $L \otimes_k K$ for arbitrary finite extensions $L, K$ of $k$ but I would settle for a reference for results such as

1) If $K / k$ is Galois with group $G$ then $K \otimes_k K \cong \oplus_{g \in G} K$.

2) If $K / k$ is purely inseparable then $K \otimes_k K$ is local with residue field $K$ and length $[K : k]$.

3) If $K / k$, $L / k$ are separable then $K \otimes_k L$ has no nilpotent elements.

Answer 1

Dear anon, the most complete reference might be Bourbaki's Algèbre, Chapter V.

For question 1, I suggest Bourbaki's Algèbre, Chapter V, §10, 4. Descente galoisienne, Corollaire . There the Master proves the more general result that the canonical morphism $$K\otimes_{k} K\to K^G: x\otimes y\mapsto (x \sigma (y))_{\sigma \in G} $$ is injective for any Galois extension $K/k$, finite or infinite, with Galois group $G$, and bijective if the extension is finite.

Question 3 is trivial from Bourbaki's point of view since for Him the definition of $K/k$ being a separable extension is that for any field extensions $L/k$, the $k$-algebra $K\otimes _{k} L$ has no nilpotents (neither $K$ nor $L$ is assumed finite-dimensional over $k$). As a concession to less enlightened mortals, He proves in §15, Exemple 3, that if the extension $K/k$ is algebraic ( for example finite-dimensional) this notion coincides with the one that you and I are familiar with: the minimal polynomial of any element in $K$ has simple roots .

For question 2, I cannot give you a reference which exactly answers your question. However a purely inseparable extension is a particular case of a primary extension and these are considered at the end of our reference, in §17,2. Produit d'extensions . The Corollaire there shows that the nilpotent radical $P$ of $K\otimes_{k} K$ is prime and since this algebra is finite-dimensional, it is local of dimension zero with unique prime ideal $P$ . We still must prove that its length is $[K:k]$. This is equivalent to the claim that the $K$-algebra $ (K\otimes_{k} K) /P $ is $K$ . This follows from the existence of the product map $K\otimes K \to K$ sending $x\otimes y$ to $xy$.The kernel of this map is exactly the unique prime ideal $P$ of $K\otimes K$ . (By the way, an excellent reference for the notion of "length" is Appendix A to Fulton's book Intersection Theory ; Example A.1.1 page 407 is relevant to the above discussion)

PS If you are not familiar with exotic languages, you will be relieved to know that this volume of Bourbaki exists in English translation.

Answer 2

I don't have references per se, but these can be proven hands-on. For (1) I would quote the normal basis theorem: if $w \in K$ is such that orbit of the Galois group on $w$ forms a $k$-basis, then by abstract nonsense, the functor $K \otimes_k -$ on $k$-algebras preserves the cokernel of the map $k[x] \to k[x]/(f) \cong K$ where

$$f(x) = \prod_{\sigma \in G} (x - \sigma(w))$$

so $K \otimes_k K \cong K[x]/(\prod_\sigma (x - \sigma(w)))$, which splits as $\prod_\sigma K[x]/(x - \sigma(w)) \cong \prod_\sigma K$ by the Chinese remainder theorem. This isomorphism is compatible with the Galois group action by the normal basis theorem.

The others can be handled by similar techniques. I think (3) actually reduces to (1) because if $E$ is a Galois extension of $k$ containing both $K$ and $L$, then $K \otimes_k L$ is a subalgebra of $E \otimes_k E$, and the latter contains no nilpotent elements by the previous calculation.

Answer 3

If $K/k$ is separable, then $K=k[\alpha]\approx k[X]/(f(X))$ where $f(X)$ is the minimal polynomial of $\alpha$. Let $L$ be a field containing $k$. Then $f(X)=f_1(X)...f_r(X)$ in $L[X]$ with the $f_i$ irreducible and distinct (because $K/k$ is separable). Therefore,

$L\otimes_kK\approx L[X]/(f(X))\approx \prod L[X]/(f_i(X))$

by the Chinese remainder theorem. This describes $K\otimes L$ completely as a product of fields when $K/k$ is separable. For example, if which $f(X)$ splits in $L$, say $f(X)=(X-\alpha_{1})\cdots(X-\alpha_{n})$, then

$L\otimes_{k}K\approx L[X]/(f(X))\approx\prod_{i}L[X]/(X-\alpha_{i})\approx\prod L_{i}$

with $L_{i}=L$. The map $L\otimes_{k}K\rightarrow L_{i}$ sends $a\otimes g(\alpha)$ to $ag(\alpha_{i})$. This takes care of 1) and 3).

As for 2), if $K=k[\alpha]$ with $\alpha^{p}\in k$, then $K\otimes_{k}K=K[\epsilon]$ where $\epsilon =\alpha\otimes1-1\otimes\alpha$ and $\epsilon^{p}=\alpha^{p}\otimes 1-1\otimes\alpha^{p}=0$. That gets you started on 2).

Answer 4

Let $\mathbb{L}/\mathbb{K}$ be an algebraic field extension (with, at least for now, no further hypotheses).

As the other answers show (in response to your first question), if $\mathbb{L}/\mathbb{K}$ is normal, separable, and finite, then we have that the explicit map $$\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle\right\rangle_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}}{\longrightarrow}\ \prod_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\mathbb{L}$$ is an isomorphism of $\mathbb{K}$-algebras. (Here the outer bracket is the correspondence from the universal property of the product $\prod_{\text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}$ and the inner bracket is the correspondence from the universal product of the coproduct $\otimes_{\mathbb{K}}$ of $\mathbb{K}$-algebras.)

As Georges mentions above, this is (by comparing dimensions as $\mathbb{K}$-vector spaces) a direct consequence of that this map is already injective as soon as $\mathbb{L}/\mathbb{K}$ is normal and separable, i.e., that under these hypotheses $$\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle\right\rangle_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}}{\longrightarrow}\ \prod_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\mathbb{L}\right)\ =\ \bigcap_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle}{\longrightarrow}\ \mathbb{L}\right)\ =\ 0$$ (as ideals of $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$). The proof is, as cited, in Bourbaki—or alternatively by the primitive element theorem or by the normal basis theorem as in the other comments.

Since this post is a useful reference, here I'd just like to present an alternative argument toward the same result. It is neither the shortest nor the most elegant, but it is (hopefully) reasonably self-contained, depends only on very basic commutative-algebraic facts, and makes no unnatural choices of basis etc.. An answer to OP's third question will be an immediate corollary.

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Before stating the main propositions, let us borrow some lemmas from the commutative algebraists. The first order of business is to represent $\mathbb{L}/\mathbb{K}$ as a filtered colimit of succesive simple extensions of $\mathbb{K}$. The "right" way to do this is to index over the comma category $\textbf{FinSet}/\mathbb{L}$, so that each object is equipped with the data of the successive simple extensions of $\mathbb{K}$ by which its value in the relevant diagram is built up. The details are straightforward but somewhat tedious, and to spare ourselves many words we'll instead resort to the more familiar construction that follows.

Denote by $$\mathcal{E}_{\mathbb{L}}$$ the filtered (by that composita of finite extensions are finite) poset category of finite subextensions of $\mathbb{L}/\mathbb{K}$ and by $$\mathscr{D}_{\mathbb{L}}\ \colon\ \mathcal{E}_{\mathbb{L}}\to\mathbb{K}\text{-alg}$$ $$\mathscr{k}_{\mathbb{L}}\ \colon\ \mathscr{D}_{\mathbb{L}} \to\ \text{const}_{\mathbb{L}/\mathbb{K}}$$ the evident diagram and (inclusion-of-subextension) cocone respectively.

Lemma 0: $\mathscr{k}_{\mathbb{L}}$ is a universal cocone of $\mathscr{D}_{\mathbb{L}}$. In particular, $$\text{colim}\left(\mathscr{D}_{\mathbb{L}}\right)\ \simeq\ \mathbb{L}/\mathbb{K}\text{.}$$

Proof of Lemma 0: The data of a cocone $\mathscr{k}'$ under $\mathscr{D}_{\mathbb{L}}$ with covertex the $\mathbb{K}$-algebra $\mathbb{A}/\mathbb{K}$ is precisely that of an element of $\mathbb{L}$ for every pair $\left(\mathbb{L}'/\mathbb{K},\alpha\in\mathbb{L}'\right)$ with $\mathbb{L}'/\mathbb{K}$ a finite subextension of $\mathbb{L}/\mathbb{K}$ such that said element of $\mathbb{L}$ depends only on $\alpha$ and that every finite tuple of such $\alpha$s jointly satisfy precisely the polynomial relations over $\mathbb{K}$ that they satisfy in $\mathbb{L}$ (as per $\mathscr{k}_{\mathbb{L}}$). That the map to this data from $\mathbb{K}$-algebra morphisms out of $\mathbb{L}/\mathbb{K}$ by pulling back along $\mathscr{k}_{\mathbb{L}}$ is a correspondence is evident. $\Box$

Lemma 1: If $\mathbb{A}$ is a commutative ring and $f\in \mathbb{A}\left[x\right]$ is a formally squarefree (i.e., the discriminant of $f$ is a unit in $\mathbb{A}$) monic polynomial, then $$\text{Nil}\left(\mathbb{A}\left[x\right]/\left(f\right)\right)\ =\ \text{Nil}\left(\mathbb{A}\right)$$ where the latter is shorthand for the ideal of $\mathbb{A}\left[x\right]/\left(f\right)$ generated by $\text{Nil}\left(\mathbb{A}\right)$. In particular, if $\mathbb{A}$ is reduced, then so is $\mathbb{A}\left[x\right]/\left(f\right)$.

Proof of Lemma 1: The claim is clear when $\mathbb{A}$ is a field (so $\mathbb{A}\left[x\right]$ a PID). The inclusion of the right hand side into the left hand side is moreover clear in general.

By the monicity of $f$, $\mathbb{A}\left[x\right]/\left(f\right)$ is free as an $\mathbb{A}$-module. It follows that \begin{align*} \text{Nil}\left(\mathbb{A}\left[x\right]/\left(f\right)\right)\ &\subseteq\ \bigcap_{\text{prime }\mathfrak{p}\ \subseteq\ \mathbb{A}} \left(\mathbb{A}\left[x\right]/\left(f\right)\overset{\text{can.}}{\longrightarrow}\mathbb{A}\left[x\right]/\left(f\right)\otimes_{\mathbb{A}}\mathbb{A}_{\mathfrak{p}}/\mathfrak{p}\right)^{\text{pre.}}\left(\text{Nil}\left(\mathbb{A}_{\mathfrak{p}}/\mathfrak{p}\left[x\right]/\left(f\right)\right)\right)\\ &=\ \bigcap_{\text{prime }\mathfrak{p}\ \subseteq\ \mathbb{A}} \text{ker}\left(\mathbb{A}\left[x\right]/\left(f\right)\overset{\text{can.}}{\longrightarrow}\mathbb{A}\left[x\right]/\left(f\right)\otimes_{\mathbb{A}}\mathbb{A}_{\mathfrak{p}}/\mathfrak{p}\right)\\ &=\ \bigcap_{\text{prime }\mathfrak{p}\ \subseteq\ \mathbb{A}}\mathfrak{p}\\ &=\ \text{Nil}\left(A\right) \end{align*} where the above are once more all ideals of $\mathbb{A}\left[x\right]/\left(f\right)$ shorthanded as generating subsets thereof and "$^{\text{pre.}}$" denotes the taking of preimages. $\Box$

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Proposition A: The evident inclusion of ideals $$\bigcap_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle}{\longrightarrow}\ \mathbb{L}\right)\ \supseteq\ \text{Nil}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\right)$$ of $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ is an equality if $\mathbb{L}/\mathbb{K}$ is normal.

Proof A: Denote by $\iota_{0},\iota_{1}\colon\mathbb{L}\to\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ the canonical $\mathbb{K}$-algebra coproduct inclusions. Given a field extension $\mathbb{M}/\mathbb{K}$ and $\mathbb{K}$-algebra morphism $$\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\overset{\varphi\ =\ \left\langle\varphi\ \circ\ \iota_{0},\ \varphi\ \circ\ \iota_{1}\right\rangle}{\longrightarrow}\mathbb{M}\text{,}$$ the normality of $\mathbb{L}/\mathbb{K}$ (and the fact that monic polynomials over $\mathbb{M}$ that split into linear factors split so uniquely) ensures that $\text{im}\left(\varphi\circ \iota_{0}\right)=\text{im}\left(\varphi\circ \iota_{1}\right)$.

We conclude that \begin{align*} \text{Nil}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\right)\ &=\ \bigcap_{\text{prime }\mathfrak{p}\ \subseteq\ \mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}}\mathfrak{p}\\ &=\ \bigcap_{\text{prime }\mathfrak{p}\ \subseteq\ \mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\overset{\text{can.}}{\longrightarrow} \left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\right)_{\mathfrak{p}}/\mathfrak{p}\right)\\ &=\ \bigcap_{\substack{\text{mor. }\varphi\ \colon\ \mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\to\mathbb{M}\text{,} \\ \mathbb{M}/\mathbb{K}\text{ a field}}}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\overset{\varphi}{\longrightarrow}\mathbb{M}\right)\\ &=\ \bigcap_{\substack{\text{mor. }\varphi\ \colon\ \mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\to\mathbb{M}\text{,} \\ \mathbb{M}/\mathbb{K}\text{ a field}}}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle 1,\ \left(\varphi\ \circ\ \iota_{0}\right)^{-1}_{\mid\ \text{im}\left(\varphi\ \circ\ \iota_{0}\right)}\ \circ\ \left(\varphi\ \circ\ \iota_{1}\right)\right\rangle}{\longrightarrow}\ \mathbb{L}\right)\\ &=\ \bigcap_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\text{ker}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle}{\longrightarrow}\ \mathbb{L}\right)\text{,} \end{align*} as claimed. (Ignoring the size issues inherent to intersecting over a technically proper-class-sized index, which are trivially rectified...) $\blacksquare$

Remark: The converse of Proposition A also holds. The proof is left as an exercise.

(Hint: As $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ is generated as a $\mathbb{K}$-algebra by elements satisfying monic polynomials over $\mathbb{K}$—namely the elements in the image of $\iota_{0},\iota_{1}$—every quotient domain of $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ is in fact a field—i.e., $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ is dimension zero. Thus it suffices to show that if $\mathbb{L}/\mathbb{K}$ is abnormal then there exists an proper ideal of $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ and an irreducible polynomial of $\mathbb{K}$ such that the third has strictly more roots in the quotient of the second by the first than it does in $\mathbb{L}$.)

Proposition B: The evident inclusion of ideals $$\text{Nil}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\right)\ \supseteq\ 0$$ of $\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}$ is an equality if $\mathbb{L}/\mathbb{K}$ is separable.

Proof B: Recall $\mathcal{E}_{\mathbb{L}}$ and $\mathscr{D}_{\mathbb{L}}$ from Lemma 0.

We first claim that $\mathbb{L}\otimes_{\mathbb{K}}\mathscr{D}_{\mathbb{L}}$ takes values in reduced $\mathbb{K}$-algebras; to see this, we strongly on $\left[\mathbb{L}':\mathbb{K}\right]$ for objects $\mathbb{L}'/\mathbb{K}$ of $\mathcal{E}_{\mathbb{L}}$ (i.e., finite subextensions of $\mathbb{L}/\mathbb{K}$). Indeed, if $\left[\mathbb{L}':\mathbb{K}\right]=1$, then $$\mathbb{L}\otimes_{\mathbb{K}}\left(\mathbb{L}'/\mathbb{K}\right)\ \simeq\ \mathbb{L}\otimes_{\mathbb{K}}\mathbb{K}\ \simeq\ \mathbb{L}\text{,}$$ which is manifestly reduced. Otherwise if $\left[\mathbb{L}':\mathbb{K}\right]>1$ then it has a proper subextension $\mathbb{L}''/\mathbb{K}$ of maximal degree over $\mathbb{K}$, and for any choice of $\alpha\in\mathbb{L}'\setminus\mathbb{L}''$, $$\mathbb{L}'\ \simeq\ \mathbb{L}''\left[x_{\alpha}\right]/\left(f_{\alpha}\right)$$ with $f$ the minimal polynomial of $\alpha$ over $\mathbb{L}''$. As $\mathbb{L}/\mathbb{K}$ is separable, this $f_{\alpha}$ is formally squarefree, so that by the inductive hypothesis and Lemma 1, $$\mathbb{L}\otimes_{\mathbb{K}}\left(\mathbb{L}'/\mathbb{K}\right)\ \simeq\ \mathbb{L}\otimes_{\mathbb{K}}\left(\mathbb{L}''\left[x_{\alpha}\right]/\left(f_{\alpha}\right)\right)\ \simeq\ \left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}''\right)\left[x_{\alpha}\right]/\left(f_{\alpha}\right)$$ is in turn reduced.

It follows (by that $\mathbb{L}\otimes_{\mathbb{K}}-$ commutes with colimits, being a coproduct) that $$\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \simeq\ \mathbb{L}\otimes_{\mathbb{K}}\text{colim}\left(\mathscr{D}_{\mathbb{L}}\right)\ \simeq\ \text{colim}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathscr{D}_{\mathbb{L}}\right)\text{,}$$ the latter a filtered colimit of reduced $\mathbb{K}$-algebras so itself reduced, as claimed. $\blacksquare$

Remark: The converse of Proposition B also holds. The proof is left as an exercise.

(Hint: Show that if $\mathbb{L}/\mathbb{K}$ is inseparable, then already $\text{Nil}\left(\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}'\right)\neq 0$ at some first stage subextension $\mathbb{L}'/\mathbb{K}$ in the colimit representation of $\mathbb{L}/\mathbb{K}$.)

Corollary: The explicit map $$\mathbb{L}\otimes_{\mathbb{K}}\mathbb{L}\ \overset{\left\langle\left\langle 1_{\mathbb{L}},\ \sigma\right\rangle\right\rangle_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}}{\longrightarrow}\ \prod_{\sigma\ \in\ \text{Aut}\left(\mathbb{L}/\mathbb{K}\right)}\mathbb{L}$$ is injective if(f) $\mathbb{L}/\mathbb{K}$ is normal and separable.

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I.e., taken together, the above propositions readily imply the claimed isomorphism. Moreover, the proof of Proposition B readily extends to showing more generally that the tensor product of a reduced $\mathbb{K}$-algebra with a separable algebraic extension of $\mathbb{K}$ is itself reduced.