# When is the tensor product of a separable field extension with itself a domain?

I'm reading Algebraic Geometry and Arithmetic Curves by Qing Liu. On page 92, in the proof of Corollary 3.2.14 d), he states that if $K \otimes_k K$ is a domain, then $K = k$. Here $K$ is a separable (not obviously finite) field extension of $k$.

Why is this true? Does it require separability?

• Sometimes any (not necessarily algebraic) field extension $k \hookrightarrow K$ is called separable iff for every field extension $k \hookrightarrow L$ the tensor product $K \otimes_k L$ is reduced. Using this definition, there are separable field extensions that are not algebraic; for instance, any purely transcendental extension is separable (see e.g. Bourbaki, Algebra, V, 15.3, Prop. 6). I guess this shows that Liu requires separable extensions to be algebraic; for otherwise the tensor product of rational function fields $k(X) \otimes_k k(Y) = k(X,Y)$ would be a counterexample. – c_c_chaos Mar 17 '15 at 21:59
• @c_c_chaos Over a year later I'm here to tell you that while it's true that $k(X) \otimes_k k(Y)$ is a domain, it is not $k(X,Y)$, and in fact it's not a field at all: There is a noninjective map $k(X) \otimes_k k(Y) \rightarrow k(T)$ which sends both X and Y to T. The map exists by universal property of the tensor product and is noninjective because X-Y is in the kernel. – user00000 Jun 6 '16 at 17:25

Separability is not necessary, but we should assume $K$ is algebraic over $k$. Let $\alpha \in K$ and suppose $\alpha \notin k$. Then it has a minimal polynomial over $k$, say $f (x)$. Then $f (x) = (x - \alpha) g (x)$ for some non-unit polynomial $g (x)$ with coefficients in $K$. But that implies that $K \otimes_k k (\alpha) \cong K [x] / (f (x))$ has a zero divisor, and $K \otimes_k k (\alpha)$ is a subring of $K \otimes_k K$, so $K \otimes_k K$ also has a zero divisor.
• What is the zero divisor in $K \otimes_k k(\alpha)$? I don't see how this comes from the factorization $f(x) = (x- \alpha) g(x)$. – Dorebell Mar 17 '15 at 20:50
• $K \otimes_k k (\alpha) \cong K [x] / (f (x))$. – Zhen Lin Mar 17 '15 at 21:31