Why is this tensor product of polynomial rings an isomorphism?

Question

Let $F \subseteq k$ be fields, $k$ algebraically closed, $I$ a radical ideal of $k[X_1, ... , X_n]$. Then $I_0 := I \cap F[X_1, ... , X_n]$ is an ideal of $F[X_1, ... , X_n]$. Suppose that $I$ can be generated by polynomials in $F[X_1, ... , X_n]$. Now the composition of $F$-algebra homomorphisms $$F[X_1, ... , X_n] \rightarrow k[X_1, ... , X_n] \rightarrow k[X_1, ... , X_n]/I$$ obviously has kernel $I_0$, so we may regard $F[X_1, ... , X_n]/I_0$ to be an $F$-subalgebra of $k[X_1, ... , X_n]/I$. In Springer's book on Algebraic Groups, he claims in this case that the natural $k$-module homomorphism (pretty sure just $k$-module, not $k$-algebra, but what do I know) $$\phi: k \otimes_F (F[X_1, ... , X_n]/I_0) \rightarrow k[X_1, ... , X_n]/I$$ is an isomorphism. Why is this? And, what can we say about $\phi$ when $I$ cannot necessarily be generated by polynomials in $F[X_1, ... , X_n]$? In any case, it looks like $\phi$ is always surjective.

Answer 1

You can use this: Why does $(A/I)\otimes_R (B/J)\cong(A\otimes_R B)/(I\otimes_R 1+1\otimes_R J)$?. In your case $R=F$, $B=F[X_1,\dots,X_n]$, $A=k$, $I=0$, and $J=I_0$. The only thing to prove is that the extension of $I_0$ to $k[X_1,\dots,X_n]$ is $I$, but this follows immediately from this answer. (In fact, if $f_1,\dots,f_t$ generate $I$, then they generate $I_0$ as well.)