# Extending an ideal of a polynomial ring to a polynomial ring with more indeterminates. Is it a tensor product?

Let $\mathbb{k}$ be a field, let $S'=\mathbb{k}[x_1,x_2,\dots,x_m]$, and let $I'\subseteq S'$ be an ideal.

For some $n>m$, let $$S=\mathbb{k}[x_1,x_2,\dots,x_n]\ \ \ (=\mathbb{k}[x_1,x_2,\dots,x_m]\otimes_\mathbb{k}\mathbb{k}[x_{m+1},\dots,x_n]),$$ and let $$I=\langle I'\rangle_{S}$$ be the $S$-ideal generated by the same generators as $I'$. That is, if $$I'=\langle f_1,\dots,f_r\rangle$$ as an ideal in $S'$, then $$I=\{s_1f_1+\cdots +s_rf_r:s_i\in S\}.$$

My question is: Is it true that $$I \cong S\otimes_{S'}I'$$ as an $S$-module?

If so, any idea on how to prove it?

Very appreciative of any help on this question. Thanks in advance.

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In this post, "ring" means "commutative ring", and "algebra" means "commutative algebra".

The following observations answer the question.

First observation. Let $R$ be a ring, let $S$ be a subring, let $I$ be an ideal of $S$, let $b$ be the $S$-bilinear map $(r,x)\mapsto rx$ from $R\times I$ to $R$, let $$\ell: R\ \underset{S}{\otimes}\ I\to R$$ be the corresponding $R$-linear map. Then

the image of $\ell$ is the ideal of $R$ generated by $I$.

This is clear. Let us denote this image by $RI$.

Second observation. Recall that an $S$-module $M$ is flat if for any $S$-linear injection $\phi:N\to P$ the $S$-linear map $$M\ \underset{S}{\otimes}\ \phi:M\ \underset{S}{\otimes}\ N\to M\ \underset{S}{\otimes}\ P$$ is injective. Then

a free module is flat.

This is almost obvious.

In the setting of the first observation, if $R$ is $S$-flat, then the natural morphism form $R\otimes_{S}I$ to $RI$ is an isomorphism.

Third observation. Let $K$ be a ring and put $$K[X]=K[x_1,\dots,x_r],$$ where the $x_i$ are indeterminates. Then

$K[X]$ is free over $K$.

Again, this is straightforward.

Fourth observation. In the above notation, set
$$K[X,Y\ ]=K[x_1,\dots,x_r,y_1,\dots,y_s],$$ where the $y_j$ are indeterminates. Then

$K[X,Y\ ]$ is free over $K[X]$.

This follows from the third observation.

Let me denote the ground field by $K$.

We have a natural surjection $$K[x_1,\dots,x_n]\otimes_{K[x_1,\dots,x_m]}I'\to K[x_1,\dots,x_n]I'.$$ Is it injective? The answer is yes. Here is the argument.

In view of standard properties of the tensor product, we have a canonical isomorphism of $K[x_{m+1},\dots,x_n]$-modules $$K[x_1,\dots,x_n]\otimes_{K[x_1,\dots,x_m]}I'\simeq K[x_{m+1},\dots,x_n]\otimes_KI',$$ and the natural morphism $$K[x_{m+1},\dots,x_n]\otimes_KI'\to K[x_1,\dots,x_n]= K[x_{m+1},\dots,x_n]\otimes_KK[x_1,\dots,x_m]$$ is clearly injective.

EDIT. More details: We have canonical isomorphisms $$K[x_1,\dots,x_n]\otimes_{K[x_1,\dots,x_m]}I'$$ $$\simeq\Big(K[x_{m+1},\dots,x_n]\otimes_KK[x_1,\dots,x_m]\Big)\otimes_{K[x_1,\dots,x_m]}I'$$ $$\simeq K[x_{m+1},\dots,x_n]\otimes_K\Big(K[x_1,\dots,x_m]\otimes_{K[x_1,\dots,x_m]}I'\Big)$$ $$\simeq K[x_{m+1},\dots,x_n]\otimes_KI'.$$

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Thanks a lot for taking the time to help me, I really appreciate it. If I understand your argument correctly, you stop where you stop because $$K[x_{m+1},\dots,x_n]\otimes_K I'\cong K[x_1,\dots,x_n]I' \ \ (=\langle I'\rangle_S)$$ as an S-module? Is this more obvious than the original isomorphism question? I'm sorry, I'm not very good with these tensor products. – Bart Patzer Jan 11 '12 at 12:08
Dear @Bart: You're welcome. I'll try to answer your comment, but I'm very busy right now. I'll do it tomorrow at the latest. By the way, +1 for your nice question! – Pierre-Yves Gaillard Jan 11 '12 at 12:47
Dear @Bart: This is just to tell you that I edited the answer. – Pierre-Yves Gaillard Jan 11 '12 at 18:58
Thanks again, this was very helpful. So let's see if I finally got it right: In order to show that $$R\oplus_SI\rightarrow RI$$ is injective, we're tensorising the exact sequence $$0\rightarrow I\rightarrow I$$ with the (in our case) flat $S$-module $R$ (over $S$), where the first $I$ is seen as an ideal of $S$, while the second (rightmost) $I$ is seen as a subset of $R$ and as an $S$-module? – Bart Patzer Jan 12 '12 at 11:57
Dear @Bart: You're welcome again. I'd say: $R$ being $S$-flat, by applying $$R\underset{S}{\otimes}\bullet$$ to the exact sequence $$0\to I\to S,$$ we get the exact sequence $$0\to R\underset{S}{\otimes}I\to R\underset{S}{\otimes}S=R.$$ Thus, the natural map $$R\underset{S}{\otimes}I\to R$$ is injective. Denoting its image by $RI$, we get $$R\underset{S}{\otimes}I\simeq RI.$$ Finally, we note that $RI$ is the ideal of $R$ generated by $I$. – Pierre-Yves Gaillard Jan 12 '12 at 13:35