Explicit description of quotient ring of $\mathbb{Z}[x]$

Question

I am studying for a prelim and I stumbled on this problem:

Describe explicitly the elements in the quotient ring $\dfrac{\mathbb{Z}[x]}{(3,x^3-x+1)}$. First of all I don't see why the ideal $(3,x^3-x+1)$ is a maximal ideal in $\mathbb{Z}[x]$. If there is anyone who can help me with this will be greatly appreciated.

Answer 1

You can solve this problem by a two step process. First, let $J = (3, x^3-x+1)$, and let $I = (3)$. These are ideals of $\mathbb Z[x]$ with $I \subset J$.

The third isomorphism theorem says that

$$\mathbb Z[x]/(3,x^3-x+1) = \mathbb Z[x]/J \cong \frac{\mathbb Z[x]/I}{J/I}$$

In other words, the ring you are looking for can be found by taking the ring $\mathbb Z[x]/I$ and modding out by an ideal therein.

Note that $\mathbb Z[x]/I \cong \mathbb F_3[x]$, where $\mathbb F_3$ is the field with three elements. Inside this ring, $J/I$ is just the ideal in $\mathbb F_3[x]$ generated by $x^3-x+1$.

The problem becomes to describe the elements of the quotient ring $\mathbb F_3[x]/(x^3-x+1)$. To do this, you should first determine whether or not $x^3-x+1$ is irreducible in $\mathbb F_3[x]$.

Answer 2

Quotients introduce relations.

In this case, we get $3=0$ and $\theta^3-\theta+1=0$. So, the quotient ring is $\mathbb Z_3[\theta]$ with $\theta^3-\theta+1=0$.

The elements are polynomial expressions in $\theta$ with coefficients in $\mathbb Z_3$.

To get unique expressions for each element $f(\theta)$, divide $f(x)$ by $x^3-x+1$ and consider the remainder $r(x)$. Then $f(\theta)=r(\theta)$. Note that $r(x)$ is $0$ or a polynomial of degree at most $2$.

Therefore, the quotient is the set $\{ a_0+a_1\theta+a_2\theta^2 : a_0,a_1,a_2 \in \mathbb Z_3\}$. The ring operations are the natural ones, subject to the basic relations $3=0$ and $\theta^3-\theta+1=0$.

Answer 3

Let $m$ be a maximal ideal of $\mathbb Z[x]$ with $(3,x^3-x+1)\subseteq m$. Assume tha $(3,x^3-x+1)\not=m$ and let $f\in m\setminus(3,x^3-x+1)\subseteq m$. Since $x^3-x+1$ is monic we have $f=g(x^3-x+1)+h$, where $g,h\in \mathbb Z[x]$ and $1\leqslant deg(h)\leqslant 2$. Now since $x^3-x+1\in m$, we have $h\in m$. We consider the following:

Case 1) $deg(h)=1$: Let $h=ax+b$. Since $3\in m$, we assume that $a, b\in\{1, 2\}$. Thus we have the following subcasis:

Subcase 1a) $a=b=1$: In this case $x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-2)=x^3-2x\in m$. Thus, $x\in m$ or $x^2-2\in m$. Hence, $1\in m$ or $x(x+2)=x^2+2=x^2-2+2(x+1)\in m$.Thus, $1\in m$ or $x\in m$ or $x+2\in m$ and so $1\in m$, a contradiction.

Other subcases are simillar.

Case 2) $deg(h)=2$: Let $h=ax^2+bx+c$. Since $3\in m$, we assume that $a, b, c\in\{1, 2\}$. Thus we have the following subcases:

Subcase 2a) $a=b=c=1$: In this case $x^2+x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-x-2)=x^3-x^2-2x\in m$. Thus, $x\in m$ or $x^2-x-2\in m$. Hence, $1\in m$ or $(x^2-x-2)-(x^2+x+1)\in m$, if $1\in m$, a contradiction. Now let $(x^2-x-2)-(x^2+x+1)\in m$, hence a polynominal of degree one is in $m$ and by case 1 we have a contradiction.

Other subcases are simillar.

Answer 4

$\dfrac{ \Bbb Z[x]}{(3,x^3-x+1)}\cong\dfrac {\Bbb Z_3[x]}{(x^3-x+1)}$ by the third isomorphism theorem.

The latter is a three dimensional vector space over $\Bbb Z_3$, with basis $\{1, \alpha, \alpha^2\}$, where $\alpha ^3-\alpha +1=0$.

So it has $27$ elements, all of the form $a\alpha^2+b\alpha +c\,,a,b,c\in\Bbb Z_3$.

Note $x^3-x+1$ is irreducible over $\Bbb Z_3$ since it doesn't have a root.