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

Proof:

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.

Thanks!

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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
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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. –  Matt N. 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. –  Matt N. Jul 27 '12 at 21:40
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2 Answers

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 $b\otimes x^k$.

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Thank you! I have to look again at this tomorrow though, it's a bit late here right now. –  Matt N. Jul 27 '12 at 21:41
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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.

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