MHB Proving Z[x] and Q[x] is not isomorphic

[cbarker1]

Dear Everyone,

What is the correct mapping between the polynomial rings $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$? The function $\phi$ is defined as $\phi(x^2+1)=\frac{1}{2}x$. I want to prove this problem by contradiction.

Thanks,
Cbarker1

 

[castor28]

Hi Cbarker1,

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.

 

[cbarker1]

I can't use the principal ideal domain. Or Euclidean domain...because I have not learn about it yet.

 

[castor28]

Hi again,

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 $\mathbb{Q}$.

 

[HOI]

The point is that because they are not isomorphic, there is no "correct mapping"! If you wanted to prove that two rings are isomorphic then you would want to find a mapping that is an isomorphism. (But there still might not be a single "correct" one.)

 

[cbarker1]

I was trying to do a contradiction proof for that problem.