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Prove that the rings $\Bbb Z[X]$ and $\Bbb Z[X,Y]$ are not isomorphic.

Can someone give me some hints on where to start ? Thanks.

Nils Matthes
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Rustin Alexandru
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3 Answers3

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There are many proofs. For example:

First proof: If $\mathbb{Z}[X] \cong \mathbb{Z}[X,Y]$, then (by modding out the ideal $(2)$) we also get $\mathbb{F}_2[X] \cong \mathbb{F}_2[X,Y]$. But $\mathbb{F}_2[X]$ is a principal ideal domain, and $\mathbb{F}_2[X,Y]$ is not.

Second proof: If $\mathbb{Z}[X] \cong \mathbb{Z}[X,Y]$, then for every commutative ring $A$ with underlying set $|A|$ we have $|A| \cong \hom(\mathbb{Z}[X],A) \cong \hom(\mathbb{Z}[X,Y],A) \cong |A|^2$. If $|A|$ is finite, this would imply $|A| = 0$ or $|A|=1$, which is absurd (take $A=\mathbb{F}_2$).

Third proof: The Krull dimension of $\mathbb{Z}[X]$ is $2$, whereas the Krull dimension of $\mathbb{Z}[X,Y]$ is $3$. But this requires some dimension theory.

In any case, you should remember the following basic technique which is used in all the proofs above (also in the answer by user129924): In order to show that two structures are not isomorphic, one tries to find an invariant which has the property that isomorphic structures have the same invariant. Then it is enough to show that the invariants differ.

Martin Brandenburg
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    +1 I like the second proof most because it just uses the universal property and voila. – Hagen von Eitzen Feb 23 '14 at 16:51
  • @Hagen: I also like it most. Actually the general case ($R[S] \cong R[T] \Rightarrow S \cong T$ for sets $S,T$) can be reduced to the case of finite $R,S,T$ by common techniques, and then one just counts. – Martin Brandenburg Feb 23 '14 at 16:55
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Their field of quotients are $\mathbb{Q}(X)$ and $\mathbb{Q}(X,Y)$ resp. so their transcendence degrees over $\mathbb{Q}$ are $1$ and $2$ respectively. Hence they can not be isomorphic.

user129924
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Here is a low-tech solution, it only uses the concept of degree.

Suppose $\varphi : \mathbb{Z}[x] \rightarrow \mathbb{Z}[x,y]$ is an isomorphism. Notice that the map is determined entirely by $\varphi(x)$. In fact, for a polynomial $f(x) \in \mathbb{Z}[x]$, we have $\varphi(f(x)) = f(\varphi(x))$ (you should show this if you haven't seen it before).

Now set $\varphi(x) = g(x,y)$, so the image of $\varphi$ is precisely the polynomials of the form $f(g(x,y))$. We need $x$ to be in the image. What does this mean about $g(x,y)$? Think about the degree of $y$ in $g(x,y)$ and in $f(g(x,y))$. Now with this conclusion and very similar reasoning, you should be able to show that $y$ can not be in the image.

RghtHndSd
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