Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to show $R[x,y]\cong R[x][y]$ using the following proposition:

Let $\varphi:R\to R'$ be a ring homomorphism. Given $\alpha_1,\alpha_2,\cdots,\alpha_n\in R'$, there exists a unique homomorphism $\Phi:R[x_1,x_2,\cdots,x_n]\to R'$ such that $$ \Phi|_{R}=\varphi, \quad \Phi(x_i)=\alpha_i,i=1,2,\cdots,n. $$

Consider the inclusion map $f:R\to R[x][y]$ which is a homomorphism. By the proposition above, there is a unique homomorphism $g:R[x,y]\to R[x][y]$ such that $$ g|_R=f $$ and $g(x)=x$, $g(y)=y$. It suffices to show that $g$ is a homomorphism. Instead of showing that $g$ is 1-1 and onto, I'm trying to construct an inverse of $g$. Consider the inclusion map $h:R[x]\to R[x,y]$. Using the proposition again, we have a unique homomorphism $l:R[x][y]\to R[x,y]$ such that $$ l_{R[x]}=h $$ and $l(y)=y$. It follows that we have a homomorphism $$ l\circ g:R[x,y]\to R[x,y] $$ such that $$ l\circ g|_{R}=id_{R[x,y]}|_R $$ and $l\circ g(x)=x$, $l\circ g(y)=y$. We also have a similar argument with $$ g\circ l: R[x][y]\to R[x][y]. $$

I've only shown that $l\circ g$ agrees with the identity map $id_{R[x,y]}$ on $R$ and $\{x,y\}$. My question:

How can I show that $l\circ g$ is the identity map $id_{R[x,y]}$? (Then similarly, I would be able to show $g\circ l=id_{R[x][y]}$).

[EDIT:] I saw a proof in Artin's Algebra, but I didn't understand how the underscored sentence works. enter image description here

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

The maps $l$ and $g$ are both ring homomorphisms, so the map $l\circ g$ is a ring homomorphism. Thus as $R[x,y]$ is generated by $R$, $x$ and $y$, and $l\circ g$ fixes these, it is the identity.

If you want to check this more carefully, you can write a general element of $R[x,y]$ in terms of $x$, $y$ and elements of $R$, apply $l\circ g$ to it, and then expand the expression using the properties of ring homomorphisms until the map $l\circ g$ is only being applied to $x$, $y$ and elements of $R$.

share|improve this answer
Hmm, this is much clearer than the proof in Artin's Algebra. –  Jack Dec 24 '12 at 21:04
Oh, I sort of like his better, having seen it - if I'd realised I'd have written my answer that way. It's a third application of the theorem you started with, this time using the "unique" part. The map $l\circ g$ extends the inclusion $R\to R[x,y]$ to a map $R[x,y]\to R[x,y]$ by mapping $x\mapsto x$ and $y\mapsto y$, but so does the identity. As this extension is unique, the two maps are the same. (This is what Artin means be "uniqueness of the substitution homomorphism".) –  Matt Pressland Dec 24 '12 at 21:07
Really this theorem is a universal property defining $R[x_1,\dotsc,x_n]$, and objects defined up to universal property are unique up to unique isomorphism. So what you're really doing is showing that $R[x,y]$ and $R[x][y]$ both have this universal property, so they must be isomorphic (and then reproducing a bit of the proof of this general fact in this case, where you show that the two maps you found are mutual inverses). Don't worry if that doesn't make sense to you, it's just some wider context. –  Matt Pressland Dec 24 '12 at 21:11
Ah, now I see how the "unique" part is applied. Thanks! What do you mean by the "universal property" defining $R[x_1,\cdots,x_n]$? –  Jack Dec 24 '12 at 21:22
The details are a little too long for a comment, but I may go back and put it into the answer, once I've thought about exactly the right way to say it. Loosely though, one way to define $R[x_1,\dotsc,x_n]$ is to say that it's the unique object such that the theorem you quoted is true. If you've ever seen the universal property of free groups, it's a similar idea to that. If you haven't, you should probably understand that example before this one, so I wouldn't worry for now. –  Matt Pressland Dec 24 '12 at 21:36
add comment

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.