3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – c_c_chaos Mar 17 '15 at 21:59
  • $\begingroup$ @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. $\endgroup$ – user00000 Jun 6 '16 at 17:25
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What is the zero divisor in $K \otimes_k k(\alpha)$? I don't see how this comes from the factorization $f(x) = (x- \alpha) g(x)$. $\endgroup$ – Dorebell Mar 17 '15 at 20:50
  • 2
    $\begingroup$ $K \otimes_k k (\alpha) \cong K [x] / (f (x))$. $\endgroup$ – Zhen Lin Mar 17 '15 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.