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It's common to think of polynomials as functions, but it was recently brought to my attention that this isn't exactly right. Consider, for example $x$ and $x^7$ in $\mathbb{Z} / 7\mathbb{Z} [x]$. These are distinct polynomials, but as functions $\mathbb{Z} / 7\mathbb{Z} \to \mathbb{Z} / 7\mathbb{Z}$ they are identical.

This made me wonder: how can we describe the kernel of the map $ \varphi : \mathbb{Z} / n\mathbb{Z} [x] \to A $ given by $\varphi(p) = (x \mapsto p(x))$ where \begin{align} A = \{p : \mathbb{Z} / n\mathbb{Z} &\to \mathbb{Z} / n\mathbb{Z} \\&\mid p \text{ is a polynomial with coefficients in } \mathbb{Z} / n\mathbb{Z}\} \end{align} (Is there a more standard notation for $A$?)

The only thing I know is that when $n$ is prime, $x^n - x \in \ker \varphi$. This is just Fermat's little theorem.

Can we give an explicit characterization of $\ker \varphi$? How should we think of elements of $\mathbb{Z} / n\mathbb{Z} [x]$ if not as functions $\mathbb{Z} / n\mathbb{Z} \to \mathbb{Z} / n\mathbb{Z}$.

Charles Hudgins
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  • Well, you should think of them as polynomials; algebraic objects in their own right with a certain universal property for the whole ring. You can think of them as "proto"-functions (or even functions), but you have to remember that when the ring is not an infinite integral domain, different polynomials may yield the same function. – Arturo Magidin Sep 04 '21 at 19:48
  • You are better off mapping to the colleciton of all functions. Given any commutative ring $R$, there is natural map $R[x]$ to $R^R={f\colon R\to R\mid f\text{ is a function }}$ taking $p(x)$ to the function mapping $a$ to $p(a)$. – Arturo Magidin Sep 04 '21 at 19:49
  • @ArturoMagidin can we give any kind of explicit characterization of the kernel of this map? – Charles Hudgins Sep 04 '21 at 19:51
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    Well, the Chinese Remainder Theorem reduces the problem to prime powers, which ought to be simpler to deal with. Off the top of my head, I know that for $n=p$ a prime the kernel is precisely the ideal generated by $x^p-x$ (two polynomials yield the same function if and only if they are congruent modulo $x^p-x$). – Arturo Magidin Sep 04 '21 at 19:54
  • @ArturoMagidin that's already a pretty interesting result. Is it elementary enough that I should attempt a proof myself, or should I look one up? – Charles Hudgins Sep 04 '21 at 19:56
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    Which? Reducing to a prime power, or the one for $p$ a prime? The latter is an easy consequence of the factor theorem, since if $f$ and $g$ give the same function, then $f-g$ is identically zero, hence must be divisible by $x-a$ for each $x\in\mathbb{Z}/p\mathbb{Z}$. – Arturo Magidin Sep 04 '21 at 19:58
  • I was asking about the latter. Someone actually just posted and then promptly deleted a post involving the factor theorem which immediately suggested the right proof. – Charles Hudgins Sep 04 '21 at 19:59
  • Yes; the factor theorem holds, but one cannot apply it inductively in general. Consider $R=\mathbb{Z}/4\mathbb{Z}$. If $p(x)\in R[x]$ has $p(1)=0$, then we know we can write $p(x) = (x-1)q(x)$. If it also has $3$ as a root, then in an integral domain we can argue that $x-3$ has to divide $q(x)$, and not just $p(x)$. But here, evaluating at $3$ gives $2q(2)$, and if $q(2)$ is even, then this is zero even if $q(2)$ is not zero. So you cannot just use the same argument for an arbitrary prime power. – Arturo Magidin Sep 04 '21 at 20:04
  • And a quick calculation shows that the polynomial $2x^2+2x$ is identically zero in $\mathbb{Z}/4\mathbb{Z}$... – Arturo Magidin Sep 04 '21 at 20:07
  • How to think of polynomials if not as functions indicate that somehow functions are simple entities that we understand well, and so it is desirable to describe other things as functions (e.g., defining a polynomial to be a function of a certain form). However, the formal set-theoretic definition of a function is deceivingly simple. Very often we go exactly the other way around: we compute $\sin(x)$ by using Taylor (ta daaaaaa) polynomials. We should be happy with polynomials as they are. That they describe functions is a feature. – Ittay Weiss Sep 04 '21 at 20:26
  • Does this answer your question? Finding the kernel of a homomorphism mapping polynomials to polynomial functions. There's also this question, which might be of interest (and which links many others in the comments) – HallaSurvivor Sep 04 '21 at 20:51
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    @HallaSurvivor I mistakenly closed this post because I thought the linked post answered my question, but it is only concerned with the integral domain case, but I am also interested in the general (say merely commutative) case. – Charles Hudgins Sep 04 '21 at 21:24
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    I voted to reopen, and that reopened it. The post that @HallaSurvivor points to is with $A$ a domain. In that case, the situation is indeed simple enough, and the kernel is trivial when $A$ is infinite, and generated by $\prod (x-a)$ as $a$ ranges over all elements of $A$ when $A$ is finite. But as even the simple case of $\mathbb{Z}/4\mathbb{Z}$ shows, the case for even the integers modulo $n$ is not that simple when $n$ is not a prime. – Arturo Magidin Sep 04 '21 at 23:09
  • @ArturoMagidin -- thanks for reopening this! OP, sorry for the screw up ^_^ – HallaSurvivor Sep 05 '21 at 01:09

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