# Nonintegral element and a homomorphism

Assume $R\subseteq S$ are rings. Choose $x\in S$ nonintegral over $R$. I want to define a homomorphism from $R[x^{-1}]$ to a field which maps $x^{-1}$ to zero.

I was trying to show that $R[x^{-1}]$ is a polynomial ring in one variable. Then I could define my map in this way $\sum_{i=0}^mr_i'(x^{-1})^i\longmapsto r_0$.

It's clear that I can show elements of $R[x^{-1}]$ like $\sum_{i=0}^nr_i(x^{-1})^i$ but at the remained part I got have a problem. I mean when I want to show that if $\sum_{i=0}^nr_i(x^{-1})^i=\sum_{i=0}^mr_i'(x^{-1})^i$ then $n=m$ and for all i, $r_i=r_i'$.

My attempt was in this way;

Let $n\geq m$, from our assumption we have $\sum_{i=0}^nr_i''(x^{-1})^i=0$ which for $0\leq i\leq m$, $r_i''=r_i-r_i'$ and for $m+1\leq i\leq n$, $r_i''=r_i$. By multiplying by $x^n$, $r_0''x^n+r_1''x^{n-1}+\cdots+r_{n-1}''x+r_n''=0$. If I were able to make this relation monic then it would be a contradiction and the problem was solved but $r_0''$ may be not invertible so what should I do now?

By the example which 'Hans' brought at below my approach was wrong.

• Hans' example shows that there is no unitary ring homomorphism from $\mathbb Z$ to any field of characteristic $\ne 2$ which sends $2$ to $0$. If you don't require the ring homomorphism to be unitary, then you have the zero homomorphism. Commented Aug 20, 2014 at 7:12

Let $x$ be nonintegral on your ring $R$ then I claim it is not in the ring $R[x^{-1}]$. Because otherwise $$x=a_nx^{-n}+\cdots+a_1x^{-1}+a_0$$ For some $a_i$s in $R$. By multiplying both sides of our equality to $x^n$ we will have $$x^{n+1}=a_n+\cdots+a_1x^{n-1}+a_0x^n$$ But it gives an integral relation for $x$ on $R$ and thus contradiction. So we showed our claim is correct.

But $x\notin R[x^{-1}]$ means $x^{-1}$ is nonunit in $R[x^{-1}]$ so there is a maximal ideal of $R[x^{-1}]$ say $\mathfrak{m}$ containing it. Denote the algebraic closure of the residual field $\dfrac{R[x^{-1}]}{\mathfrak{m}}$ by $k$. And define the ring homomorphism $\phi$ be composition of the following two ring homomorphisms. $$R[x^{-1}]\overset{\pi}{\longrightarrow}\dfrac{R[x^{-1}]}{\mathfrak{m}}\overset{j}{\longrightarrow}k$$ Which the first one is the canonical map and the second one is inclusion.

$\phi$ is a ring homomorphism from $R[x^{-1}]$ to an algebraic closed field which its kernel is exactly $\mathfrak{m}$ and thus nontrivial.

• @user26857:I checked the proof of Mr.AmirHosein step by step and I'm sure it's extremely true
– M.H
Commented Sep 9, 2014 at 15:46
• @MaisamHedyelloo This answer has nothing to do with the original answer, that is, before the massive edit. Moreover, even the question was edited to fit the above answer, which I find rather unprofessional. (Btw, I've also checked the proof and I can say that the field $k$ doesn't necessarily contain $R$. A simple example is $R=\mathbb Z/4\mathbb Z$ and the polynomial ring $S=(\mathbb Z/4\mathbb Z)[x]$. If $\mathfrak m=(2,x^{-1})$, then $\mathfrak m\cap\mathbb Z/4\mathbb Z\ne 0$.) Commented Sep 9, 2014 at 18:01
• @user26857 thank you for warning me that the part 'containing R' is not the case.
– H.W.
Commented Sep 10, 2014 at 14:46

Let $R=\mathbb{Z}$ and $S=\mathbb{Q}$. Then $x=\frac{1}{2} \in \mathbb{Q}$ is not integral over $\mathbb{Z}$ since $\mathbb{Z}$ is integrally closed. But of course $\mathbb{Z}=\mathbb{Z}[x^{-1}]$ which is not the polynomial ring. I guess you need some additional assumptions.