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Let $d$ be a square free integer and $p$ an odd prime, prove that if $\mathbb{F}_p[X]/(X^2-d)$ has a nilpotent element, then $X^2-d$ is a square in $\mathbb{F}_p[X]$.

It should be an equivalence, the other implication is very simple to prove. However, this implication is causing me a lot of problems.

Ok so through the relation that $X^2=d$ in $\mathbb{F}_p[X]/(X^2-d)$, we have that every element can be written as $aX+b$, with $a,b \in F_p$. The fact that $\mathbb{F}_p[X]/(X^2-d)$ contains a nilpotent element implies the existence of $aX+b$ and $n$ integer such that $aX+b\neq 0$, i.e. $aX+b \notin (X^2-d)$ but $(aX+b)^n\in (X^2-d)$.

How can I use this to deduce that $(X^2-d)$ is a square?

cnymfais
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  • Hint: to contain is to divide. Since $(aX + b)^n \in (X^2 - d)$, then $(aX+b)^n = q(X) (X^2 - d)$ for some $q(X) \in \mathbb{F}_p[X]$. By unique factorization and degree considerations, then $X^2 - d = u(aX+b)^2$ for some $u \in \mathbb{F}_p^\times$. – Viktor Vaughn Jun 06 '24 at 18:22
  • I answered below. Meanwhile though, I am confused. Can't you simply use the fact that $d$ not a square $\implies$ $X^2-d$ irreducible in $\mathbb{F}_p$ $\implies$ $\mathbb{F}_p[X]/(X^2-d)$ a field? Then no nilpotent elements follow right away. – Mike Jun 06 '24 at 18:47
  • Please answer this comment OP as this does require clarification. – Mike Jun 06 '24 at 18:51
  • I edited my answer, I apologize for any confuiusion – Mike Jun 06 '24 at 19:14

2 Answers2

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We have three possibilities:

  1. If $x^2-d$ is irreducible in $\mathbb{F}_p[x]$, then $\mathbb{F}_p[x]/(x^2-d)$ is a quotient by a maximal ideal, hence a field, so it contains no irreducibles. The fact that $(x^2-d)$ is a maximal ideal follows because $\mathbb{F}_p[x]$ is a PID, so irreducibles generatee maximal ideals.

  2. If $x^2-d$ is reducible in $\mathbb{F}_p[x]$, then it factors. Say it factors as $(x-\alpha)(x-\beta)$. If $\alpha\neq\beta$, then by the Chinese Remainder Theorem we have $$\frac{\mathbb{F}_p[x]}{(x^2-d)} = \frac{\mathbb{F}_p[x]}{(x-\alpha)}\times\frac{\mathbb{F}_p[x]}{(x-\beta)} \cong \mathbb{F}_p\times\mathbb{F}_p.$$ The last isomorphism via "evaluate at $\alpha$" and "evaluate at $\beta$". This ring does not have nonzero nilpotent elements.

  3. If $x^2-d$ is a perfect square in $\mathbb{F}_p[x]$, $x^2-d = (x-\alpha)^2$. Then $\mathbb{F}_p[x]/(x^2-d)$ has the nonzero nilpotent element $x-\alpha$.

Thus we see that $\mathbb{F}_p[x]/(x^2-d)$ has nilpotents if and only if $x^2-d$ is a perfect square in $\mathbb{F}_p[x]$. Note that it doesn't matter whether $d$ is squarefree or not: it really only matters if $d$ has zero, one, or two square roots in $\mathbb{F}_p$. In fact, note that we can only have $x^2-d$ be a perfect square in two situations: if $p=2$, or if $d=0$. In particular, under your assumptions (that $d$ is a squarefree integer and $p$ is odd), $\mathbb{F}_p[x]/(x^2-d)$ never has nonzero nilpotents.

Indeed, if $p=2$ and there is a square root $\alpha$ for $d$, then $(x-\alpha)^2=x^2-\alpha^2 = x^2-d$. If $d=0$, then we have $x^2$.

Conversely, if $d\neq 0$ and $p\neq 2$, then either $d$ has not square roots and $x^2-d$ is irreducible; or $\alpha$ is a square root for $d\neq 0$, hence $\alpha\neq 0$, and so the polynomial factors as $(x-\alpha)(x+\alpha)$; but $\alpha\neq-\alpha$, since $p\neq 2$, so $x^2-d$ is not a perfect square in $\mathbb{F}_p[x]$.

Arturo Magidin
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  • But, doesn't $x^2-d$ not irreducible $\implies$ $x^2-d$ factoring into $(x-\alpha)(x-\beta)$; $\alpha, \beta \in \mathbb{F}_p$ imply e.g. $\alpha \in \mathbb{F}_p$ is a root of the polynomial $X^2-d=0$ and thus $\alpha^2 =d$ and thus $d$ a square in $\mathbb{F}_p$? What am I missing here – Mike Jun 06 '24 at 19:32
  • @Mike Not sure why that matters? $\alpha^2=d$ in $\mathbb{F}_p$, but the assumption isn't "$d$ is not a square in $\mathbb{F}_p$", the assumption is "$d$ is a squarefree integer". – Arturo Magidin Jun 06 '24 at 19:35
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This holds much more generally, and the general proof is more enlightening no more difficult, viz.

Lemma $K[x]/f\ \text{has a $\:\!\rm\color{#0a0}{nontrivial}\ \color{#c00}{nilpotent}$}\!\iff\! f\ \text{not squarefree}\,$ for $\,f,g\!\in\! K[x],\,$ field $K$

If $f\,$ is not squarefee then $\,f = g^2h,\,$ nonunit $\,g,\,$ so $\,\color{#c00}{(gh)^2\!=0}\color{#0a0}{\neq gh}\,$ in $K[x]/f,\ $ else

if $\:\!f\,$ is squarefree $\:g^n\! =\! 0\,$ in $K[x]/f\! \iff\!\!$ $ f\mid g^n\!\!\!\!\underset{\rm\color{#0af}{S}\!\!}\iff\! f\mid g\!\iff\! g\!=\!0\,$ in $K[x]/f\,$ by $\rm\color{#0af}{S}$=here

Bill Dubuque
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