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I am wondering if the following is true:

Let $A, R$ commutative rings, $\varphi:R\rightarrow A$ a surjective homomorphism, see $A$ as an $R$-algebra. Then $$ R[X_1,\dots,X_m]\otimes_R A\approx A[X_1,\dots,X_m]$$ I am thinking in applying this to the case of $R\rightarrow R/J$ where $J$ is an ideal of $R$. Any help would be greatly appreciated.

user26857
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mathma
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1 Answers1

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It is important to know that your special case is the only case, up to isomorphism.

Here is a proof: define $\phi_1:A\to A[x_1, \dots x_n]$ by $a\mapsto a$ and $\phi_2:R[x_1, \dots x_n]\to A[x_1, \dots x_n]$ by $\sum_i a_Ix^I\mapsto \sum_I \varphi(a_I)x^I$. These maps both coincide on $R$, so by the universal property of tensor products, we have a map $\phi:A\otimes_R R[x_1, \dots x_n]\to A[x_1, \dots x_n]$. There is also a map, $f:A[x_1, \dots x_n]\to A\otimes_R R[x_1, \dots x_n]$, by $f(a_Ix^I)=a_I\otimes x^I$. It is easy to check that the composition of these two are the identity. This proves the statement.

Pax
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