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. 
 A: 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 
$$
K[x_1,\dots,x_n]\otimes_{K[x_1,\dots,x_m]}I' 
$$
$$
\simeq\Big(K[x_{m+1},\dots,x_n]\otimes_KK[x_1,\dots,x_m]\Big)\otimes_{K[x_1,\dots,x_m]}I'
$$
$$
\simeq K[x_{m+1},\dots,x_n]\otimes_K\Big(K[x_1,\dots,x_m]\otimes_{K[x_1,\dots,x_m]}I'\Big)
$$
$$
\simeq K[x_{m+1},\dots,x_n]\otimes_KI'.
$$
