Show that $\mathbb{Z}[\pi]$ is isomorphic to $\mathbb{Z}[x]$ Assuming $\pi$ is not the root of any integer polynomial, show that $\mathbb{Z}[\pi]$ is isomorphic to $\mathbb{Z}[x]$.
Is it valid to say that since $\pi$ is transcendental and since $\pi$ is not an integer that $\mathbb{Z}[\pi]$ will not add anything to $\mathbb{Z}[x]$? How can I say this more formally?
 A: To expand on your comment in Robert's hint/answer.
Yes, define $\varphi : \mathbb{Z}[x] \to \mathbb{R}$ by $\varphi(f(x))=f(\pi)$.
This is an "evaluation homomorphism". Obviously, $\varphi(f+g)=\varphi(f)+\varphi(g)$, $\varphi(fg)=\varphi(f)\varphi(g)$, $\varphi(1)=1$ (it is a homomorphism).
Notice that (essentially by definition) the image of $\varphi$ is $\mathbb{Z}[\pi]$.
Finally, for the kernel, suppose that $\varphi(f(x))=0$. This means that $f(\pi)=0$. But we know that $\pi$ is transcendental and so is not the root of any nonzero polynomial with integer coefficients. Therefore, $f(x)=0$. This shows the kernel is trivial.
Therefore, the domain is isomorphic to the image and you get $\mathbb{Z}[x] \cong \mathbb{Z}[\pi]$. 
There's nothing really special about the integers and $\pi$ here. In general, you have $R[x] \cong R[\alpha]$ if $\alpha$ transcends $R$. In fact, that's essentially the definition of transcendental.
A: Hint: consider the obvious mapping of $\mathbb Z[x]$ to $\mathbb Z[\pi]$.
