Tensor product of purely inseparable extension 
I would like to understand the tensor product of $A=\Bbb F_2(\sqrt{t})\otimes_{\Bbb F_2(t)}\Bbb F_2(\sqrt{t})$.

The extension $L/k:=\Bbb F_2(\sqrt{t})/\Bbb F_2(t)$ is a finite extension of degree $2$ purely inseparable since the minimal of $\sqrt{t}$ over $k$ is $P=X^2-t$ and any element $(a+b\sqrt{t})/(c+d\sqrt{t})\in L$ has its square in $k$.
It must have nilpotents base on readings of different threads but if I compute the tensor product $L\otimes_kL$, with $L\cong k[X]/(X^2-t)$, I have:
$$A=L\otimes_kL=L\otimes k[X]/(X^2-t)=L[X]/(X^2-t)=L[X]/(X-\sqrt{t})^2$$
(edited: can't use CRT here!)
How could I find the nilpotents?
 A: $A$ does have nilpotents, namely $X-\sqrt{t}$ by Frobenius (there is nothing mysterious here; this is simply what you have already observed). The equivalence you posted previously, namely $A = L[X]/(X-\sqrt{t})^{2} \cong L[X]/(X-t) \times L[X]/(X-t)$, was not correct; we cannot apply the Chinese remainder theorem here. 
To compute nilradical $R$ of $A = L[X]/\langle (X-\sqrt{t})^{2} \rangle$, let $I = \langle (X-\sqrt{t})^{2} \rangle$ and recall that $R$ is the intersection of all prime ideals of $A$. The prime ideals of $A$ are in bijective correspondence with the prime ideals of $L[X]$ containing $I$ via the canonical quotient homomorphism $\pi \colon L[X] \to A$. It thus suffices to compute the intersection $J$ of all prime ideals of $L[X]$ containing $I$; then $\pi(J) = R$. Any prime ideal $\mathfrak{p}$ in $L[X]$ which contains $I$ must contain $X-\sqrt{t}$, hence $\langle X - \sqrt{t} \rangle$. On the other hand, $\langle X - \sqrt{t} \rangle$ is a prime ideal of $L[X]$ containing $I$, so we see $J = \langle X - \sqrt{t} \rangle$. 
