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

Cbarker1

Active member
Jan 8, 2013
236
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

Well-known member
MHB Math Scholar
Oct 18, 2017
245
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

Active member
Jan 8, 2013
236
I can't use the principal ideal domain. Or Euclidean domain....because I have not learn about it yet.
 

castor28

Well-known member
MHB Math Scholar
Oct 18, 2017
245
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 $\mathbb{Q}$.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
384
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

Active member
Jan 8, 2013
236
I was trying to do a contradiction proof for that problem.