What is $\mathbb{F}_p(\sqrt{X})\otimes_{\mathbb{F}_p(X)} \mathbb{F}_p(\sqrt{X})$? I am trying to understand what $\mathbb{F}_p(\sqrt{X})\otimes_{\mathbb{F}_p (X)} \mathbb{F}_p(\sqrt{X})$ is. $\mathbb{F}_p(\sqrt{X})$ is the field of rational functions in $\sqrt{X}$. What is it isomorphic to? Is it a field?
I noticed that $\mathbb{F}_p (X)$ is a subfield of $\mathbb{F}_p (\sqrt X)$. But I am not very sure how to proceed from here.
 A: This is somewhat similar to one of your previous questions. Note that $\mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}(X)[T]/\langle T^{2}- X\rangle$. This polynomial is irreducible by Eisenstein's criterion, and is separable if $p \neq 2$. In this case, we have
$$\mathbb{F}_{p}(\sqrt{X}) \otimes_{\mathbb{F}_{p}(X)} \mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}(\sqrt{X}) \otimes_{\mathbb{F}_{p}(X)} \mathbb{F}_{p}(X)[T]/\langle T^{2} -X \rangle \cong \mathbb{F}_{p}(\sqrt{X})[T]/\langle T^{2}-X\rangle$$
As noted above, if $p \neq 2$, then $T^{2}-X$ factors as $(T+\sqrt{X})(T-\sqrt{X})$ over $\mathbb{F}_{p}(\sqrt{X})$, so by the Chinese Remainder Theorem, 
$$\mathbb{F}_{p}(\sqrt{X})[T]/\langle T^{2}-X\rangle = \mathbb{F}_{p}(\sqrt{X})[T]/\langle (T-\sqrt{X})(T+\sqrt{X}) \rangle \cong  \\
\mathbb{F}_{p}(\sqrt{X})[T]/\langle (T-\sqrt{X}) \rangle \times \mathbb{F}_{p}(\sqrt{X})[T]/\langle (T+\sqrt{X}) \rangle \cong \mathbb{F}_{p}(\sqrt{X}) \times \mathbb{F}_{p}(\sqrt{X})$$
On the other hand, if $p = 2$, then $T^{2}-X$ factors as $(T-\sqrt{X})^{2}$, so $\mathbb{F}_{p}(\sqrt{X})$ is not a separable extension of $\mathbb{F}_{p}(X)$. In this case, I don't know what progress one can make in simplifying the tensor product in question - it may not have a nice description.
