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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.

1 Answers 1

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

New answer

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.

Old answer

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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    There,$I$finally got it! This was most kind of you. Have a nice day!2012-01-12