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- Thread starter Cbarker1
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- Oct 18, 2017

- 245

I don't see what is the point of your function $\phi$: it is not even defined on the whole of $\mathbb{Z}$.

In reference to the title of you post, to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic, you could use the fact that $\mathbb{Q}[x]$ is a Euclidean domain, and therefore a principal ideal domain.

On the other hand, $\mathbb{Z}[x]$ is not a principal ideal domain; for example, in $\mathbb{Z}[x]$, the ideal $I =\langle x,2\rangle$ is not principal.

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- Oct 18, 2017

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In fact, it is even simpler. If $\theta:\mathbb{Z}[x]\to\mathbb{Q}[x]$ is an isomorphism, then $\theta(1) = 1$, because any ring homomorphism must map $1$ to $1$.

Now, in $\mathbb{Q}[x]$, we have $1 = \dfrac12+\dfrac12$. If $f(x)=\theta^{-1}(\dfrac12)$, we must have $f(x)+f(x) = 1$ in $\mathbb{Z}[x]$, and it is obvious that no integer polynomial satisfies that relation.

We have simply elaborated on the fact that $\mathbb{Z}$ is not isomorphic to $\mathbb{Q}$.

- Jan 30, 2018

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