Subring $R \subset \Bbb{Q}[X]$. Evaluation hom. $R/(f) \to \Bbb{Z}/(f(a))$ gives me a weird result (not true), but everything looks correct.

Question

Let $R = \text{Int}[X] \subset \Bbb{Q}[X]$ be the subring of integer-valued polynomials.

Fix an $a \in \Bbb{Z}$.

Let $f \in R$ be one such irreducible, monic polynomial. Then $\psi: R/(f) \to \Bbb{Z}/(f(a)), \ \psi(g + (f)) = g(a) + (f(a))$ is a surjective ring homomorphism. If $g(X) = h(X) \pmod f$ then $f(X) \mid g(X) - h(X) \implies f(a) \mid g(a) - h(a)$.

$\psi(h + g) = h(a) + g(a) + (f(a)) = \psi(h) + \psi(g)$. Similarly $\psi(hg) = \psi(h)\psi(g)$.

$\psi$ is surjective since we can achieve any $z \in \Bbb{Z}$ by evaluating integer polynomials $X + d$ over all $d \in \Bbb{Z}$ which are in $R$.

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But $f \in R \subset \Bbb{Q}[X]$ so if $f$ is irreducible in $\Bbb{Q}[X]$ then it is also irreducible in $R$, so that $(f)$ (in $R$) is also maximal in $R$ and $R/(f)$ is a field.

Then $\psi$ is actually an isomorphism since it's kernel has to be $(0)$. Thus $R/(f) \simeq \Bbb{Z}/(f(a))$ for any $a \in \Bbb{Z}$, so you get weird isomorphisms for all the different moduli $f(a)$.

Therefore, something in my understanding is not correct. Can you help me suss this out?

Answer 1

There are a couple of problems: your assertion that $R/(f)$ is a field is unwarranted. Your assertion that $\psi$ is one-to-one is unwarranted.

1. You try to deduce that $R/(f)$ is a field from the conclusion that $f$ is irreducible in the case at hand ($f$ monic and irreducible in $\mathbb{Q}[x]$). But the implication $$f\text{ is irreducible}\implies (f)\text{ is maximal}$$ holds in a PID, but not in an arbitrary domain. You can only conclude that $(f)$ is maximal among principal ideals. For instance, $2$ is irreducible in $\mathbb{Z}[x]$, but $(2)$ is not maximal: it is properly contained in $(2,x)$. Likewise, here $(f)$ is generated by an irreducible, but unless $R$ is a PID (and I’m pretty sure it is not), then you have no warrant to assert that $R/(f)$ is a field.

2. Your assertion that $\psi$ is an isomorphism of fields assumes that (1) the target is a field; and (2) $\psi$ is one to one. It’s possible that the target is trivial (if $f(a)=\pm 1$); in which case you don’t get an isomorphism of fields. And, more relevant, $\psi$ need not be one-to-one: in general, its kernel will contains $(x-a)+(f(x))$. If we are assuming that $f(x)$ is irreducible in $\mathbb{Q}[x]$ and monic, then $x-a\in(f(x))$ only if $x-a=f(x)$, which will not be the case in most instances. So $\psi$ usually has non-trivial kernel, which is okay since $(f)$ is only maximal among principal ideals, but not necessarily all ideals of $R$.