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I am self-studying ring theory. Come to the following question,

For which values of $p$ the rings $\mathbb{Z}_p[x]/(x^3-1)$ and $\mathbb{Z}_p[x]/((x-2)(x-3)(x-5))$ are isomorphic?

In general when $R[x]/(f(x))$ and $R[x]/((g(x))$ are isomorphic for a commutative ring $R$ with unity?

Here is my try after seeing Dietrich Burde's comment:

$\mathbb{Z}_p[x]/((x-2)(x-3)(x-5)) = \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p$ by CRT. $x^3-1 = (x-1)(x^2+x+1)$. Now does $(x-1)$ and $(x^2+x+1)$ are comaximal ideal in $\mathbb{Z}_p[x]$?

One thing that is clear to me is that for $p=3$ they are not isomorphic because one has a non-zero nilpotent element and the other does not.

Thank you.

user26857
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KAK
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  • Please avoid saying "I have no idea how to start". This sounds like you didn't try anything. You could start with the Chinese remainder theorem, and trying for which $p$ the polynomial $x^3-1$ splits into three linear factors. – Dietrich Burde Jun 01 '23 at 08:30

1 Answers1

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Assume first of all that $p \geqslant 5$. In this case, $2 \neq 3$, $2 \neq 5$ and $3 \neq 5$ so you can use the CRT as you did (be careful, it is not the case for $p = 2$ or $3$ !).

If $x^2 + x + 1$ factors as $(x - a)(x - b)$ with $a \neq b$, $a \neq 1$, $b \neq 1$ in $\mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$, you have by the CRT, $$ \mathbb{Z}_p[x]/(x^3 - 1) = \mathbb{Z}_p[x]/((x - 1)(x - a)(x - b)) = \mathbb{F}_p^3 = \mathbb{Z}_p/((x - 2)(x - 3)(x - 5)). $$ Notice that $1$ is not a root of $x^2 + x + 1$ since we assumed that $p \geqslant 5$. And if $a = b$, $(x - a)^2 = x^2 - 2ax + a^2 = x^2 + x + 1$ implies that $-2a = 1$ so $a = -2^{-1}$ (exists because $p \neq 2$) and $a^2 = 1 = 2^{-2}$ thus $4 = 1$ so $p = 3$, which contradits $p \geqslant 5$.

It proves that $x^2 + x + 1$ has two disctint roots, both distinct to $1$ if and only if it has a root. In the case where it doesn't have, you have that $$ \mathbb{Z}_p[x]/(x^3 - 1) = \mathbb{Z}_p[x]/((x - 1)(x^2 + x + 1)) = \mathbb{F}_p \times \mathbb{F}_{p^2} \neq \mathbb{F}_p^3, $$ because $x - 1$ and $x^2 + x + 1$ are coprime.

In the case $p \geqslant 5$, you deduce that your rings are equal if and only if $\exists x \in \mathbb{Z}_p, x^2 + x + 1 = 0$, if and only if $\exists x \in \mathbb{Z}, p|x^2 + x + 1$.

Edit : I corrected some mistakes in this part

The cases $p = 2$ and $3$ are a bit special because $$ \mathbb{Z}_2[x]/((x - 2)(x - 3)(x - 5)) = \mathbb{Z}_2[x]/(x(x - 3)^2) = \mathbb{F}_2 \times \mathbb{F}_2[x]/(x^2), $$ and $$ \mathbb{Z}_3[x]/((x - 2)(x - 3)(x - 5)) = \mathbb{Z}_3[x]/(x(x - 2)^2) = \mathbb{F}_3 \times \mathbb{F}_3[x]/(x^2). $$ Notice that, $$ \mathbb{Z}_2[x]/(x^3 - 1) = \mathbb{Z}_2[x]/((x - 1)(x^2 + x + 1)) = \mathbb{F}_2 \times \mathbb{F}_4 \neq \mathbb{F}_2 \times \mathbb{F}_2[x]/(x^2), $$ because $x^2 + x + 1$ has no root in $\mathbb{Z}_2$ and, $$ \mathbb{Z}_3[x]/(x^3 - 1) = \mathbb{Z}_3[x]/((x - 1)^3) = \mathbb{F}_3[x]/(x^3) \neq \mathbb{F}_3 \times \mathbb{F}_3[x]/(x^2), $$ so the answer is no for $p = 2$ and for $p = 3$.

For general $f$ and $g$ in $\mathbb{Z}_p$, I think (I am not sure) that you can use the same method to prove that $\mathbb{Z}_p[x]/(f) = \mathbb{Z}_p[x]/(g)$ if an only if you can write $f = \prod_{k = 1}^n f_k$ and $g = \prod_{k = 1}^n g_k$ with for all $k$, $f_k$ and $g_k$ irreducible of the same degree. This is due to the fact that when $f$ is irreducible, $\mathbb{Z}_p[x]/(f) = \mathbb{F}_{p^{\mathrm{deg}(f)}}$ only depends on the degree of $f$.

For a general $R$, I don't think there is a method, it is actually pretty hard to tell even for $R = \mathbb{Q}$.

Cactus
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  • Beside being incomplete (it is not stated clearly for each primes $p$ the polynomial $X^2+X+1$ is irreducile modulo $p$), this answer still has some mistakes (for instance, for $p=3$). Moreover, the question is about isomorphisms, not about equalities. So one should provide reasons for why the objects are not isomorphic. (I'll downvote the answer in order to draw the attention to other readers to be careful.) – user26857 Jun 01 '23 at 18:52
  • I corrected a typo for $p = 3$. When I write $=$, it means "isomorphic", and I don't thnik there is a method to know whether $p|x^2 + x + 1$ for some $x$. We know that there exists an $x$ such that $p|x^2 + 1$ if and only if $p \equiv 1 [4]$, so maybe we can find a similar result with $x^2 + x + 1$. I'll try to look for an answer but I don't have it right now. – Cactus Jun 02 '23 at 06:26
  • When claim that two rings are not isomorphic an argument should follow, isn't it? – user26857 Jun 02 '23 at 21:32
  • The proof is let to the reader xD. No, seriously, $\mathbb{F}p \times \mathbb{F}{p^2} \neq \mathbb{F}p^3$ because in $\mathbb{F}_p^3$, $a = (1,0,0)$, $b = (0,1,0)$ and $c = (0,0,1)$ verify $a \neq 0$, $b \neq 0$, $c \neq 0$ and $ab = ac = bc = 0$ but in $\mathbb{F}_p \times \mathbb{F}{p^2}$, if $a = (a_1,a_2)$, $b = (b_1,b_2)$ and $c = (c_1,c_2)$ verify the above assumptions, $a_1b_1 = a_1c_1 = b_1c_1 = 0$ so at least two of the $x_1$ are zero. Assume WLOG that $a_1 = b_1 = 0$. And $a_2b_2 = 0$ which contradicts $a \neq 0$ and $b \neq 0$. – Cactus Jun 04 '23 at 09:08
  • $\mathbb{F}_2 \times \mathbb{F}_4 \neq\mathbb{F}_2 \times \mathbb{F}_2[x]/(x^2)$ because the second one contains a non-zero nilpotent element : $(0,x)$ but the first one doesn't.

    And $\mathbb{F}_3[x]/(x^3) \neq \mathbb{F}_3 \times \mathbb{F}_3[x]/(x^2)$ because the first one contains a nilpotent element of order $3$ : $x$ but the first one doesn't because if $(a,f(x)\ [x^2])^3 = (a^3,f(x)^3\ [x^2]) = (0,0)$, then $a^3 = 0$ so $a = 0$ and $x^2|f(x)^3$ hence $x|f(x)^3$ hence $x|f(x)$ (because $x$ is prime). We deduce that $(a,f(x)\ [x])^2 = (0,0)$.

     – Cactus Jun 04 '23 at 09:12
  • All these comments fit into the answer. This link https://math.stackexchange.com/questions/1311702, too. – user26857 Jun 04 '23 at 16:38