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I'm looking for the necessary and sufficient conditions for any prime $p$ and any positive integer $n$ to make the quotient rings $\mathbb{Z}_p[x]/(x^n+1)$ and $\mathbb{Z}_p[x]/(x^n-1)$ not only rings but fields.

I know that the quotient ring $\mathbb{Z}_p[x]/f(x)$ is a field if and only if $f(x)$ is irreducible over $\mathbb{Z}_p$:

  1. Then $\mathbb{Z}_p[x]/(x^n-1)$ can never be a field, as $1$ is a root of $x^n-1$ in $\mathbb{Z}_p[x]$, making it reducible, am I right?

  2. For $\mathbb{Z}_p[x]/(x^n+1)$:

    2.1. If $n\geq p$, then $f(x)=x^n+1$ might have a root in $\mathbb{Z}_p$, as one of the two following cases happens: either (a) all integers in $\mathbb{Z}_p^*$ are roots of $f(x)$, or (b) there is a monic polynomial $g(x)$ such that $g(x)\equiv 0 (\text{mod }p)$ with $\text{deg}(g)<p$ and the solutions of $g(x)\equiv 0(\text{mod }p)$ are the solutions of $f(x)\equiv 0 (\text{mod }p)$. Note: In (b), if the equation $g(x)\equiv 0(\text{mod }p)$ has no solutions, then $f(x)\equiv 0(\text{mod }p)$ has no solutions as well.

    2.2. If $n<p$, then for all integers $a$ with $\gcd(a,p)=1$, we have that $x^n\equiv a(\text{mod }p)$ has exactly $\gcd(n,p-1)$ solutions if $$a^{\frac{p-1}{\gcd(n,p-1)}}\equiv 1(\text{mod }p),$$ or no solution at all if $$a^{\frac{p-1}{\gcd(n,p-1)}}\not\equiv 1(\text{mod }p).$$ Then by taking $a=p-1$, we have that $x^n+1\equiv 0(\text{mod }p)$ which is equivalent to $x^n\equiv p-1(\text{mod }p)$ has exactly $\gcd(n,p-1)$ solutions if $$(-1)^{\frac{p-1}{\gcd(n,p-1)}}\equiv 1(\text{mod }p),$$ or no solutions at all if $$(-1)^{\frac{p-1}{\gcd(n,p-1)}}\not\equiv 1(\text{mod }p).$$

My questions are:

  1. Just to be sure, $\mathbb{Z}_p[x]/(x^n-1)$ is never a field, right?
  2. Is there any easier criteria or direct theorem I'm missing for the case $\mathbb{Z}_p[x]/(x^n+1)$ which could give me a more straight-forward condition for $p$ and $n$?

Thank you.

liwuen
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    $x^n-1$ is irreducible for $n=1$. Don't forget the trivial cases, because they will be sad and feel left out. – Arthur Jun 04 '20 at 21:44
  • If $n=pm$ then $x^n+1=(x^m+1)^p$. This leaves the case $n$ not a multiple of $p$. – lhf Jun 04 '20 at 22:00
  • If $n=pm$ then $n\geq p$ which falls under my case 2.1. – liwuen Jun 04 '20 at 22:03
  • Your reasoning in 2.1 isn't quite right. For example, $x^4+1$ has no roots in $\mathbb{Z}/3\mathbb{Z}$. Recall from Fermat's little theorem that $a^{p-1}\equiv 1\ (\text{mod}\ p)$ for any $a\in\mathbb{Z}/p\mathbb{Z}$, so the question of whether $x^n+1$ has a root is equivalent to the question of whether $x^m+1$ has a root where $n\equiv m\ (\text{mod}\ p-1)$ and $m\in{1,,2,\ldots,p-1}$. – Anonymous Jun 04 '20 at 22:17
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    You are right. After checking the theorem where I based my conclusion again, it only says that if $g(x)\equiv 0(\text{mod }p)$ has solutions, then $f(x)\equiv 0(\text{mod }p)$ also has. This does not justify that $g(x)\equiv 0(\text{mod }p)$ will have any solution at all. – liwuen Jun 04 '20 at 22:34

2 Answers2

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$1$ is a root of $x^n-1$ thus it is irreducible iff $n=1$.

$x^n+1$ is irreducible iff $f=[\Bbb{F}_p(\zeta_{2n}):\Bbb{F}_p]=n$ iff the least integer such that $2n | p^m-1$ is $m=n$.

ie. $f=order(p\bmod 2n) $. It divides $\varphi(2n)$.

$n| \varphi(2n)$ implies $n=2^k$.

And $order(p\bmod 2^{k+1})=2^k$ implies that $k=0$ or $1$

and hence either $n=1$ or $n=2$ and $p\equiv 3\bmod 4$.

reuns
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  • For what I can see cyclotomic extensions are quite useful. I will take a deep look at them. Thank you! – liwuen Jun 04 '20 at 23:20
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First consider $x^n-1$ . . .

If $n=1$ then $x^n-1=x-1$ which is irreducible in $\mathbb{Z}_p[x]$.

If $n > 1$ then $x^n-1$ has the factor $x-1$, so $x^n-1$ is reducible in $\mathbb{Z}_p[x]$.

Next consider $x^n+1$ . . .

If $n=1$ then $x^n+1=x+1$ which is irreducible in $\mathbb{Z}_p[x]$.

If $n$ is odd and $n > 1$, then $x^n+1$ has the factor $x+1$, so $x^n+1$ is reducible in $\mathbb{Z}_p[x]$.

If $n$ is an even positive integer but not a power of $2$, then $n$ has an odd prime factor, $q$ say, with $q < n$. Letting $m=n/q$, it follows that $x^n+1$ has the factor $x^m+1$, so $x^n+1$ is reducible in $\mathbb{Z}_p[x]$.

If $n=2$ then $x^n+1=x^2+1$ which factors mod $p$ if and only if $-1$ is a quadratic residue mod $p$, which happens if and only if $p=2$ or $p\equiv 1\;(\text{mod}\;4)$.

If $n=4$ then $x^n+1=x^4+1$ which is reducible mod $p$ for all primes $p$.

$\qquad$Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime

If $n$ is a power of $2$ with $n > 4$, then since $x^4+1$ is reducible mod $p$ for all primes $p$, it follows that $x^n+1$ is reducible mod $p$ for all primes $p$.

quasi
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  • Thank you! I should had split the values of $n$ and $p$ as you did. I think that if I did the split for 2.2, then I would arrive to your result. Thank you! – liwuen Jun 04 '20 at 23:16