# 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?

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What part are you finding it hard to prove? – Chris Eagle Jul 12 '12 at 21:29
Please share your failed attempts! – wildildildlife Jul 12 '12 at 21:52
My attempt at a proof began by defining the map which sends $x$ to $x-c$ and is the identity on R. Then I tried to show that the map was a homomorphism, and the calculations, which involved lots of summations and binomial coefficients and whatnot, seemed intractable to me. I rarely see such awful proofs. Usually there is a much cleaner way to prove anything. – PrimeRibeyeDeal Jul 12 '12 at 22:55
@PrimeRibeyeDeal For the proof, see my answer. – M Turgeon Jul 12 '12 at 22:58
@MTurgeon Yes, thanks. I am currently reading a proof of the theorem you gave. – PrimeRibeyeDeal Jul 12 '12 at 23:07

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, $X\mapsto X+c$).

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Thanks. This is essentially what I was looking for: a general fact I can cart around in my head, and one that brings me closer to understanding the "algebraic relations" that polynomials embody. – PrimeRibeyeDeal Jul 12 '12 at 23:38

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 $$\phi(f+g)=\sum_{n=0}^\infty (a_n+b_n)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n+\sum_{n=0}^\infty a_n(x-c)^n=\phi(f)+\phi(g)$$ $$\phi(fg)=\sum_{n=0}^\infty\left( \sum_{k=0}^n a_kb_{n-k}\right)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)\phi(g)$$

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 $\,\phi(1)=1\,$ .

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Thanks, I very much appreciate you writing this out explicitly. I perpetrated most of the same calculations when I originally embarked, but I fell apart at the middle equality in the line where you show the preservation of multiplicative structure. Perhaps I could see the implication on a day my brain feels more energetic, but is there some way you could make it slightly easier to see? – PrimeRibeyeDeal Jul 12 '12 at 23:35
Yes @PrimeRibeyeDeal: write it down carefully! For example, let's do the first ones: $$\sum_{n=0}^\infty\left(\sum_{k=0}^n a_nb_{n-k}\right)(x-c)^n=a_0b_0+(a_0b_1+a_1b_0)(x-c)+(a_0b_2+a_1b_1+a_2b_0)(x-c)‌​^2+...$$and now write down the other expression separately – DonAntonio Jul 12 '12 at 23:57
Ok, thank you very much. – PrimeRibeyeDeal Jul 13 '12 at 14:18

Hint: The map $x\to x+c$ is his inverse.

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And one needs to show that these maps preserve both addition and multiplication, etc. – Lubin Jul 12 '12 at 21:35