Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a subring of a commutative unital ring $B$.

Can you tell me if my proof of the following claim is correct?

Claim: $B \otimes_A A[X] \cong B[X]$


It's enough to show that $B[X]$ satisfies the universal property of $B \otimes_A A[X]$, that is if $N$ is any $R$-module and $b^\prime: B \times A[X] \to N$ any bilinear map then there exists a unique linear map $l: B[X] \to N$ such that $l \circ b = b^\prime$.

Define $b: B \times A[X] \to B[X]$ as $(r,p(x)) \mapsto rp(x)$. This is bilinear. Now define $l: B[X] \to N$ as $l: p(x) \mapsto b^\prime((1,p(x)))$. It remains to be shown that this $l$ is unique and linear. Linearity directly follows from the bilinearity of $b^\prime$. So let $l^\prime: B[X] \to N$ be another linear map such that $l^\prime \circ b = b^\prime$. Then $$ l(p(x)) = b^\prime ((1,p(x)) = l^\prime \circ b ((1,p(x)) = l^\prime(p(x))$$ hence $l$ is unique and $B \otimes_A A[X] \cong B[X]$ follows.


share|cite|improve this question
Showing that $B[X]$ has the universal property of the $A$-module $B\otimes_AA[X]$ would only give you an $A$-module isomorphism $B[X]\cong B\otimes_AA[X]$. But in fact $B[X]$ and $B\otimes_AA[X]$ are $B$-algebras, not just $A$-modules, and the natural isomorphism between them is a $B$-algebra isomorphism. – Keenan Kidwell Jul 27 '12 at 18:42
I think it is good habit to always state in which category your 'isomorphisms' should hold, otherwise $\cong$ is ambiguous. – wildildildlife Jul 27 '12 at 21:22
@wildildildlife Good point. Thanks. – Rudy the Reindeer Jul 27 '12 at 21:39
@KeenanKidwell Thank you! Unfortunately it's a bit late here so I will have to read your comment again tomorrow. – Rudy the Reindeer Jul 27 '12 at 21:40
up vote 4 down vote accepted

You have a problem in that $p(x)\in B[x]$ may not be in $A[x]$, so defining the map by setting $p(x)\mapsto b'((1,p(x))$ is not valid.

A more direct route might be to use the map you have from $B\otimes _AA[x]\to B[x]$ and provide an inverse; for example, the map that sends $bx^k$ to $b\otimes x^k$.

share|cite|improve this answer
Thank you! I have to look again at this tomorrow though, it's a bit late here right now. – Rudy the Reindeer Jul 27 '12 at 21:41

By the universal property of $(B \otimes_A A[x],b)$ there exists a unique linear map $l: B \otimes A[x] \to B[x]$ such that for the map $b^\prime : B \times A[x], (b_0, p(x) ) \mapsto b_0 p(x)$ we have $l \circ b = b^\prime$.

We claim that $l$ has a two-sided inverse: define $\varphi: B[x] \to B \otimes A[x]$, $\sum b_i x^i \mapsto \sum b_i \otimes x^i$. Then $$ l(\varphi ( \sum b_i x^i)) = l(\sum b_i \otimes x^i) = \sum b^\prime (b_i, x^i) = \sum b_i x^i$$ and $$ \varphi (l ( b_0 \otimes p(x))) = \varphi (b^\prime (b_0, p(x)) = \varphi (b_0 p(x)) = b_0 \otimes p(x)$$ which proves the claim.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.