$a^n=0\,\Rightarrow\, X-a\mid X^n$ (nilpotent Factor Theorem)

Question

Proposition 1.6 in A. Kleiman A Term of Commutative Algebra says that if $R$ is a ring and $a \in R$, then the map $\pi:R[X]\rightarrow R$ given by $\pi(X) = a$ has kernel generated by $(X-a)$ and $R[X]/(X-a)$ is isomorphic to $R$. But if $a$ is nilpotent, say $a^n = 0$, then $\pi(X^n) = a^n = 0$ so $X^n \in \operatorname{Ker}(\pi)$, but $X^n \notin \langle X-a\rangle$ (the ideal generated by $X-a$).

Am I missing something? Certainly, the proposition is true if $R$ is an integral domain.

TeX note: please use $\operatorname{Ker}$ \operatorname{Ker} (or \DeclareMathOperator\Ker{Ker} and then just \Ker if it's something you'll use frequently) in place of just $Ker$, and $\langle X - a\rangle$ \langle X - a\rangle in place of $<X - a>$. For the latter, note the difference in spacing between $X^n \notin \langle X - a\rangle$ (X^n \notin \langle X - a\rangle)—correct spacing, although the statement is incorrect, as @DaveBenson explained—and $X^n \notin <X - a>$ X^n \notin <X - a>. I have edited accordingly. — LSpice, Oct 26 '23 at 19:33
@LSpice to begin with, this is not research level maths, and so it is not suitable on MathOverflow... — Vladimir Dotsenko, Oct 26 '23 at 19:40
@VladimirDotsenko, re, that is true but, I think, orthogonal to my comment. (I did not answer the question.) Hopefully this will be migrated to MathSE, and, when it is there, it will still be good for the TeX to be correct. — LSpice, Oct 26 '23 at 20:39
@LSpice what I meant is that the expectation of LaTeX proficiency on MO and MSE is clearly very different, and so I was confused by the fact that it were the TeXnical matters that you prioritised... — Vladimir Dotsenko, Oct 27 '23 at 04:15

score 13 · Accepted Answer · answered Oct 26 '23 at 19:24

13

Yes, you're missing something. It's that $$X^n=X^n-a^n=(X-a)(X^{n-1}+aX^{n-2}+\dots+a^{n-2}X+a^{n-1})$$ is in the ideal generated by $X-a$.

answered Oct 26 '23 at 19:24

Dave Benson

406

Thank you. Duh! :) How did I miss that? – rwilsker Oct 26 '23 at 21:55
1

I was wondering the same thing, but didn't like to mention it out of politeness. At least the question was properly migrated, and belongs quite comfortably here. – Dave Benson Oct 26 '23 at 22:01
@rwilsker We don't need the quotient: $,a,$ is a root of $,x^n\Rightarrow x-a\mid x^n,$ in $R[x],$ by the Factor Theorem. – Bill Dubuque Oct 26 '23 at 23:00

Bill Dubuque · Answer 2 · 2024-02-23T15:12:15.180

$\,a^n\!=\!0\,\Rightarrow\, a\,$ is a root of $\,x^n\Rightarrow x-a\mid x^n\,$ in $R[x]\:\!$ by FT= Factor Theorem. $\ \bf\small QED$

Remark $ $ This method is worth explicit emphasis since it seems many students overlook that FT applies here. Generally FT $\Rightarrow x-a\mid f(x)-f(a)\,$ (we can compute the quotient from that if need be as in Dave's answer, but we don't need it here). This is also closely related to inverting a unit + nilpotent (by simpler multiples - analogous to rationalizing a denominator), see here.

$a^n=0\,\Rightarrow\, X-a\mid X^n$ (nilpotent Factor Theorem)

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