How to see that the shift $x \mapsto (x-c)$ is an automorphism of $R[x]$? In the process of studying irreducibility of polynomials, I encountered the criterion that $p(x)$ is irreducible if and only if $p(x-c)$ is irreducible. When trying to determine what properties of the ring were preserved under this map $x \mapsto (x-c)$, which appears sometimes to be called the shift isomorphism, I read that it was an isomorphism of the polynomial ring $R[x]$, but my attempts to prove that fact only led me through some difficult calculations, at which I generally fail.
So how does one prove that the map is an isomorphism of $R[x]$? Is it an isomorphism for all rings $R$ and for any number of variables?
Is this just a specific case of a more general phenomenon?
 A: Hmmm...what about directly? 
$$f(x)=\sum_{n=0}^\infty a_nx^n\,,\,g(x)=\sum_{n=0}^\infty b_nx^n\,\in R[x]$$
with $\,\,a_n=b_n=0 \,\,\text{ for all but a finite number of indices}$ . Since 
$$f(x)+g(x)=\sum_{n=0}^\infty (a_n+b_n)x^n\,\,,\,f(x)g(x)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)x^n$$
we get that, denoting the map by$\,\phi(x):=x-c\,$, we have
$$\phi(f+g)=\sum_{n=0}^\infty (a_n+b_n)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n+\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)+\phi(g)$$
$$\phi(fg)=\sum_{n=0}^\infty\left( \sum_{k=0}^n a_kb_{n-k}\right)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)\phi(g)$$
Remember: all the above are in fact finite sums, so we can re-arrange them as we wish. Also, $\,\phi(r)=r\,\,\,\forall\,r\in R\,$, so in particular $\,\phi(1)=1\,$ .
A: As Lubin noted in a comment above, the first thing to check is that the map which sends $p(x)$ to $p(x-c)$ is indeed a ring homomorphism. To do this, we have the following tool:
Theorem: Let $R$ be a commutative ring. The polynomial ring $R[X]$ satisfies the following universal property: given a commutative ring $A$ containing $R$ and an element $a$ of $A$, there is a unique ring homomorphism 
$$\phi:R[X] \to A$$ such that $\phi(X)=a$. 
(Note that there exists a similar version for the polynomial ring in finitely many variables.) Hence, by this universal property there exists a unique ring homomorphism that sends $X$ to $X-c$, for any $c\in R$. And as azarel noted above, this is in fact a ring isomorphism, since you have an inverse (namely, $X\mapsto X+c$).
A: Hint: The map $x\to x+c$ is his inverse.
