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I was sitting in the Math room at my school and was reading the AMA Monthly and came across the proof for the following problem:

Let $R$ be any UFD that is not a field. Suppose that $R$ has only finitely many units. If $f(x) \in R[x]$ is any nonconstant polynomial, then $f(a)$ is composite for some $a \in R$.

They prove this as follows:

Suppose that $R$ has $k$ units and let $f(x)$ have degree $d$. Choose any distinct $kd+1$ elements of $R$. Then for one of these elements $b$, $f(b)$ is not a unit, as otherwise $f(x)$ would have to take the same value more than $d$ times and so would be constant. Let $p$ be any prime dividing $f(b)$. Choose any distinct $(k+1)d+1$ elements of $R$ congruent to $b$ mod $p$. Then for one of these elements $a$, $f(a)$ is neither a unit multiple of $p$ nor is equal to $0$, then $f(a)$ is composite.

My questions are as follows:

$1$. I understand in both selections we are applying some sort of Pigeonhole Principle, but in the first case why would it be $kd+1$ elements? Wouldn't it be sufficient to look at $k+1$ elements of $R$? Or would this merely work for 'most' polynomials in $R[x]$? I fail to see why we choose that many again in the $(k+1)d+1$ choice but I imagine it's for about the same reason.

$2$. Why are we guaranteed to find elements among those $(k+1)d+1$ elements equivalent to $b$ mod $p$? I don't see why that should always be the case.

$3$. The author never address the case when $R$ may have infinitely many units. Certainly, there are uncountably infinite rings with countably infinite units. Then the proof does not work. Or are we to merely take the result for granted since no polynomial can take on prime values infinitely often?

Bill Dubuque
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Kyle L
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1 Answers1

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You may find this clearer. First, $R$ is infinite, since finite domains are fields. Since the set of units $U$ is finite, $f$ takes some value $\,f(b) = \color{#c00}{a\not\in U\cup \{0\}}\,$ (else $f$ takes a values $\,v\in U\cup \{0\}$ infinitely many times, so $f-v$ has infinitely many roots, contra a nonzero polynomial over a domain has no more roots than its degree). Similarly $f(ax+b)$ takes some value $\,f(ar+b)\,$ not in $\,\color{blue}{aU\cup\{0\}}$. Note $\,a\mid f(ar+b)\,$ since $\,{\rm mod}\ a\!:\ f(ar+b)\equiv f(b) = a\equiv 0,\,$ hence $\,f(ar+b) = ac\,$ for $\,c\in R.\,$ By construction, $\,\color{#0a0}{c\not\in U\cup\{0\}}$ (else $\,f(ar+b) = ac\in \,\color{blue}{aU\cup\{0\}}).\,$ Therefore, since both $\,\color{#c00}a\,$ and $\,\color{#0a0}c\,$ are nonzero nonunits, we infer $\,ac = f(ar+b)\,$ is a composite value of $f$.

Bill Dubuque
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