# Tensor product and compositum of fields

Let E/k, F/k be two arbitrary field extensions of k. My question is:

1. Is there a field extension M/k s.t. E/k, F/k are subextensions of M/k? Alternatively, can we talk about compositum fields without assuming a larger field?

2. If the answer to the above question is yes, can we construct such a M/k explicitly (by tensor product, direct product, localization, quotient etc.)?

3. Is the k-algebra $E\otimes_k F$ never the zero ring?

-
The tensor product of two nonzero vector spaces is nonzero. – Pierre-Yves Gaillard Aug 11 '11 at 9:35
The compositum always exists. See for instance Algèbre, Bourbaki, Section V.2.4, Cor. to Prop. 4. – Pierre-Yves Gaillard Aug 11 '11 at 9:51

As you seem to have realized the concept of a compositum is a bit troubling, if we are not working inside a bigger field. Any way, here come a few quick and dirty answers:

1+2) Yes (up to identification). The tensor product $E\otimes_k F$ is a commutative $k$-algebra. If $I$ is a maximal ideal in there (call upon Zorn's lemma to get the existence), then $M=E\otimes_k F/I$ is a field. The mappings $e\mapsto e\otimes 1 +I$ and $f\mapsto 1\otimes f +I$ are then homomorphisms of $k$-algebras, and thus injective (scores of details to check here). Therefore we can identify their images with $E$ and $F$ respectively. However, the choice of $I$ may make quite a difference. For example, if $E$ and $F$ are isomorphic (think: $\mathbf{Q}(\root 3\of 2)$ and $\mathbf{Q}(\omega\root3\of2)$, $\omega=(-1+i\sqrt3)/2$), then their images are equal for an appropriate choice of $I$, but may intersect trivially for another one.

[Edit: It is possible that the images of $E$ and $F$ never intersect trivially. A trivial intersection occurs in my example case, but clearly not always. Sorry about any possible confusion this error in the original version may have created.]

3) No. This never happens. The tensor product of two non-trivial vector spaces is never zero. You get a basis for the tensor product from pairwise elementary tensors of basis elements of the factor spaces. When we mod out a maximal ideal as above, some linear dependencies may or may not be introduced (the ideal may be zero).

-
Congratulations on the 10k :-) – Asaf Karagila Nov 13 '11 at 22:23

Consider the tensor product $A=E\otimes _kF$.

a) The $k$-algebra $A$ is nonzero because any choice of bases $(e_i)_{i \in I}$ of $E$ and $(f_j)_{j \in J}$ of $F$ will yield a basis $(e_i \otimes f_j)_{(i,j) \in I\times J}$ of $A=E\otimes _kF$ ( This answers your question 3.)

b) The set $Spec(A)$ of prime ideals of the non-zero ring $A$ is thus non empty.
Consider a prime ideal $P \subset A$, the quotient $A/P$ and the fraction field $k(P)=Frac(A/P)$.
The compositon of $k$-morphisms $E \to A=E\otimes _k F \to A/P \to k(P)$ is necessarily injective (since $E$ is a field!) and exhibits $k(P)$ as an extension of $E$. Similarly $k(P)$ is canonically an extension of $F$. And finally $k(P)$ is a compositum of $E$ and $F$. (This answers your question 2. and part of your question 1. For the rest of 1. see c) below)

c) Every compositum of the extensions $E,F$ of $k$ is obtained (up to isomorphism) by the procedure above applied to a suitable prime $P\in Spec(A)$. Moreover different primes yield non-isomorphic extensions. So $Spec(A) =Spec(E\otimes _kF)$ exactly classifies the isomorphism classes of composita of $E$ and $F$. In particular, since $Spec(A)$ has in general more than one element, it does not make sense to talk of the compositum of $E$ and $F$ in the abstract, that is if $E$ and $F$ are not given as subfields of some extension of $k$. (This answers the rest of your question 1.)

d) A paraphrase of the above is that we have described the set underlying the affine scheme $Spec(E\otimes _k F)$. More generally, the description in c) is the key point in the description of the set underlying the fiber product $X\times _S Y$ of two arbitrary schemes $X,Y$ over an arbitrary scheme $S$.

Given two field extensions $k\to E, k\to F$, let me explain in elementary terms (that is without tensor products) what a compositum of these is.
It is the data of a field extension $k \to K$ and of a pair of $k$-morphisms ($E\to K, F\to K$), subject to the condition that the union of the images of $E$ and $F$ in $K$ generate (in the field sense !) the field $K$. An isomorphism of two such extensions is an isomorphism $K\to K'$of $k$-extensions making obvious diagrams commute.
The key point to keep in mind is that the extension $k\to K$ alone does not determine the compositum.
This is perfectly illustrated by Matt's and Pierre-Yves's exchange in the comments where ( a little confusingly!) the three extensions $E,F,K$ of $k=\mathbb R$ are the inclusion $k=\mathbb R \to K=\mathbb C$ and they have two composita with the $same$ extension $k\to K$ which are nevertheless non-isomorphic: this is possible because the composita have different pairs of $k$-morphisms ($E\to K, F\to K$) in their data.
To see if I got it right: If you take $k=\mathbb R$, $E=F=\mathbb C$, you get precisely two non-isomorphic composita. – Pierre-Yves Gaillard Aug 11 '11 at 12:09
@Pierre: Dear Pierre-Yves, Yes, but you have to interpret the concept of isomorphism appropriately: abstractly, the two composita are both isomorphic to $\mathbb C$, but they are realized as composita in two different ways: in the first way $E$ and $F$ are both mapped to $\mathbb C$ identically; in the second way, $E$ (say) is mapped to $\mathbb C$ identically, but $F$ is mapped via complex conjugation. Regards, – Matt E Aug 11 '11 at 12:49
Dear @Matt: Thanks for your very clear explanation! Say that $C$ is a compositum of $E$ and $F$ over $k$. Viewing $C$ as a compositum means viewing it as a $E\otimes_k F$-algebra, whereas viewing it as an extension of $k$ means viewing it as a $k$-algebra. Right? – Pierre-Yves Gaillard Aug 11 '11 at 12:59