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Let $\Bbb Z/p$ be the finite field with $p$ elements.

Consider $\Bbb Z/p[X]$, the ring of polynomials with coefficients in $\Bbb Z/p$.

Consider also the ring $P(\Bbb Z/p)$ of all polynomial functions on $\Bbb Z/p$.

Let $\varphi$ be the morphism $\Bbb Z/p[X]\to P(\Bbb Z/p)$, linking to each polynomial its polynomial function.

Is it correct that the kernel of $\varphi$ is the ideal $(X^p-X)$?

InvisiblePanda
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Michiel Vermeulen
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    Yes. ${}{}{}{}$ – Qiaochu Yuan Dec 31 '12 at 07:58
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    I've noticed on this site that many people use the notation $\mathbb Z/p$ instead of $\mathbb Z/p\mathbb Z$. Is this a kind of new trend in algebra? Which textbooks are using this notation? (I'm simply curious to see where is this coming from.) –  Dec 31 '12 at 11:39
  • And the notation $\mathbb Z_p$ for $\mathbb Z/p\mathbb Z$ is even worst ! –  Dec 31 '12 at 21:42

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This shouldn't be terrible to see: $k[X]$ is always a PID (for $k$ a field), and $X^p - X = \prod_{a \in \mathbb{Z}_p} (X - a)$ is defo in the kernel by little fermat. So if $(X^p - X)$ weren't the whole kernel, it would be generated by some $f$ properly dividing this guy, if $f$ omits a given factor $(X - a_0)$, then boop, $f$ doesn't vanish as a function on $a_0$!

uncookedfalcon
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  • Thank you! And what if we replace the field Z/p by the ring Z/n? What is the kernel of the morphism from Z/n[X] to P(Z/n) linking to each polynomial it's polynomial function? Note that Z/n[X] is not a PID. – Michiel Vermeulen Jan 04 '13 at 00:43