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I know they are both UFDS, but not Euclidean domains nor PIDS. The argument that I have seen showing that they are not isomorphic goes along the lines of saying the number of invertible elements in each ring is different. If this valid, then I still have trouble seeing why the order of the invertible subset should be preserved under ring isomorphisms.

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One idea would be to look at the units in in $\mathbb{Z}[x]$ and units in $\mathbb{Q}[x,y]$ –  user9413 Apr 10 '12 at 16:30
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Every property of a ring is preserved under ring isomorphisms; that's the point of ring isomorphisms. Explicitly, if $\phi : A \to B$ is a ring isomorphism, then $a$ invertible means $1 = \phi(1) = \phi(a a^{-1}) = \phi(a) \phi(a^{-1})$, so $\phi(a)$ also has an inverse, and vice versa. –  Qiaochu Yuan Apr 10 '12 at 16:31
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Another argument, that doesn't explicitly mention inverses: if $\phi$ is an isomorphism from $\mathbb{Q}[x,y]$ to $\mathbb{Z}[x]$, then $\phi(1/2)$ is an element $f$ of $\mathbb{Z}[x]$ such that $f+f=1$. –  Chris Eagle Apr 10 '12 at 16:34
    
Right. Chris correctly points out that the two aren't even isomorphic as abelian groups under addition. –  Qiaochu Yuan Apr 10 '12 at 16:50
    
Minor nitpick: every property of a ring that is derived from the ring structure is preserved by isomorphisms. There are other properties not preserved by ring isomorphisms, such as the set-theoretic identity of its elements (e.g. the property "is the empty set a member of the underlying set of the ring?"). –  Hurkyl Apr 10 '12 at 17:05
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Suppose that $R$ and $S$ are rings, and that $f:R\to S$ is an isomorphism. Let $R^\times\subset R$ and $S^\times \subset S$ be the respective subsets of invertible elements.

Your goal is to show that $f(R^\times)=S^\times$, which (because $f$ is a bijection) demonstrates that $|R^\times|=|S^\times|$. A good way to proceed would be to show that $f(R^\times)\subseteq S^\times$ and that $S^\times\subseteq f(R^\times)$. Hint: Use the fact that $f$ and $f^{-1}$ are homomorphisms.

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If $\phi:\mathbb{Z}[x]\rightarrow \mathbb{Q}[x,y]$ is a ring homomorphism, then $\phi(1)=1$. It then follows that for all $n\in \mathbb{Z}$, $\phi(n)=n$ (if you're not convinced of this, it would be worth sitting down and writing out a proof).

Assume that $\phi$ is injective. Pick $n\in \mathbb{N}$, and suppose that $\frac{1}{n}\in im(\phi)$. Then there exists $f(x)\in \mathbb{Z}[x]$ with $\phi(f)=\frac{1}{n}$. Then $\phi(nf)=1=\phi(1)$, implying that $nf(x)\equiv 1$ and hence $n=\pm 1$. Thus, $\frac{1}{n}\notin im(\phi)$ for any $n\geq 2$.

Therefore no injective homomorphism from $\mathbb{Z}[x]$ to $\mathbb{Q}[x,y]$ can be onto, and hence $\mathbb{Z}[x] \ncong \mathbb{Q}[x,y]$.

Alternatively: Suppose again that $\phi:\mathbb{Z}[x]\rightarrow \mathbb{Q}[x,y]$ is injective, and let $\phi(x)=g(x,y)$. So, for any $$f(x)=a_0+a_1x+a_2 x^2 + \cdots + a_m x^m\in \mathbb{Z}[x],$$ we have $$\phi(f)=a_0+a_1g(x,y)+a_2(g(x,y))^2 + \cdots + a_m (g(x,y))^m\in \mathbb{Q}[x,y].$$

Now, consider $g(x,y)$. If $g(x,y)$ is a constant, then $\phi$ is certainly not onto (and hence not an isomorphism), so assume that $deg(g)=k>0$. Let $qx^ay^b$ (with $q\in \mathbb{Q}$ and $a,b\in \mathbb{N}$ with $a+b=k$) be a term of $g$ having maximal degree. Then $g(x,y)-qx^ay^b\notin im(\phi)$ (it would be a good exercise to write down a short proof explaining why). Therefore once again, $\phi$ cannot be onto.

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