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.


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.)

  • $\begingroup$ Let's assume $I = f_1 k[X_i] + \cdots + f_t k[X_i]$ so that $I_0 = f_1 F[X_i] + \cdots + f_t F[X_i]$. Then $$C := I \otimes_F 1 + 1 \otimes_F I_0 = 0 + 1 \otimes_F I_0 $$ where $1 \otimes_F I_0$ is the abelian group generated by $1 \otimes g$, where $g \in I_0$. Equivalently, $1 \otimes_F I_0$ is the $F$-module generated by $1 \otimes f_1, ... , 1 \otimes f_t$. My question is, why does $C$ correspond to $I$ under the isomorphism $k \otimes_F F[X_1, ... , X_n] \rightarrow k[X_1, ... , X_n]$? The scalars in $C$ are not allowed to be in $k$, right? $\endgroup$ – D_S Sep 30 '15 at 14:39
  • $\begingroup$ @D_S $1\otimes f_i$ are in fact the generators of $C$ as an ideal. (Here all are ring isomorphisms.) $1\otimes_FI_0$ is also an ideal. $\endgroup$ – user26857 Sep 30 '15 at 14:44
  • $\begingroup$ Oh you're right. I misstated the result I was thinking of from Hungerford. $\endgroup$ – D_S Sep 30 '15 at 14:53

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