# Ideal generated by two elements is maximal in $\mathbb{Z}[x]$

The question is to show that $$I = (x^4 + x^3 + x^2 + x + 1, 2) \subset \mathbb{Z}[x]$$ is a maximal ideal.

I'm familiar with the results that $$R / I$$ is a field iff $$I$$ is maximal, and $$R/I$$ is a field iff $$I = (p(x))$$ where $$p(x)$$ is an irreducible polynomial.

I'm a little thrown off by the ideal being generated by two elements. I know that $$x^4 + x^3 + x^2 + x + 1$$ is irreducible (shifting $$x$$ to $$x+1$$ and then applying Eisenstein's criterion) and $$2$$ is irreducible, so does that mean the ideal $$I$$ generated by both $$2, x^4+x^3+x^2+x+1$$ is irreducible (and thus prime because $$\mathbb{Z}[x]$$ is a UFD), so $$R / I$$ is a field, and then $$I$$ is maximal?

• How about $J = (x^4+x^3+x^2+x+1,2,x^2) \supsetneq I$. We have $x \notin J$ so $J \neq \mathbb{Z}[x]$. – mechanodroid Feb 17 at 21:03
• @mechanodroid How about $x+1=x^4+x^3+x^2+x+1-x^2(x^2+x+1)\in J$ and $x=x(x+1)-x^2\in J$? In fact, $J=\mathbb Z[x]$. – user26857 Feb 17 at 23:18
• @user26857 Thanks, I knew I was missing something obvious. – mechanodroid Feb 17 at 23:21
• A principal ideal can not be maximal in $\mathbb Z[x]$. – user26857 Feb 18 at 14:23
• This holds for irreducible polynomials $p$ in $K[x]$, with $K$ a field. – user26857 Feb 18 at 14:56

## 3 Answers

I'm not sure what you mean when you ask whether $$I$$ is irreducible. It's probably helpful to use the third (and I guess also the second) isomorphism theorem, which will tell you: $$\mathbb{Z}[x]/(x^4+x^3+x^2+x+1,2) \cong \mathbb{F}_2[x]/(x^4+x^3+x^2+x+1).$$ Now if you show that the polynomial $$x^4+x^3+x^2+x+1$$ is irreducible over $$\mathbb{F}_2$$, then since $$\mathbb{F}_2[x]$$ is a UFD, this will give you a field.

Let $$p (x)=x^4+x^3+x^2+x+1$$. Then $$\frac {\Bbb Z [x]}{\langle2,p (x)\rangle}\cong\frac {\Bbb F_2[x]}{\langle p (x)\rangle}$$ where $$\Bbb F_2$$ denote the field with two elements. Since $$p$$ is irreducible in $$\Bbb F_2$$ the ring above is a field, hence $$I$$ is maximal.

Here's an elementary proof. Let $$J \subseteq \mathbb{Z}[x]$$ be an ideal such that $$I \subsetneq J$$. We claim that $$J = \mathbb{Z}[x]$$.

Let $$f(x) = a_nx^n + \cdots + a_1x + a_0 \in J \setminus I$$. If $$\deg f = n \ge 4$$, we can subtract a multiple of $$x^4 + x^3 + x^2+ x + 1$$ to reduce the degree of $$f$$, namely $$f(x) - a_nx^{n-4}(x^4 + x^3 + x^2+ x + 1) \in J$$ and has degree $$\le n-1$$.

Hence without loss of generality we can assume that $$\deg f \le 3$$ so $$f(x) = a_3x^3+a_2x^2+a_1x+a_0$$. Furthermore, by subtracting a multiple of $$2$$, we can reduce the coefficients $$a_0, a_1, a_2, a_3$$ to $$0$$ or $$1$$.

The only nontrivial possibilities for $$f$$ are $$x,x+1,x^2,x^2+1,x^2+x,x^2+x+1,x^3,x^3+1, x^3+x$$$$x^3+x+1,x^3+x^2,x^3+x^2+1,x^3+x^2+x,x^3+x^2+x+1$$

You can fiddle with these directly to show that $$1 \in J$$.

If $$f(x) = x$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - x(x^3+x^2+x+1)\in J$$ If $$f(x) = x+1$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - (x+1)(x^3+x)\in J$$ If $$f(x) = x^2$$, we have $$1 = 1-x^5 + x^5 = (x^4 + x^3 + x^2+ x + 1)(-x+1) - x^2\cdot x^3 \in J$$ If $$f(x) = x^2+1$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - (x^2+1)(x^2+x)\in J$$ If $$f(x) = x^2+x$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - (x^2+x)(x^2+1)\in J$$ If $$f(x) = x^2+x+1$$, we have $$1 = -(x^4 + x^3 + x^2+ x + 1)x - (x^2+x+1)(x^3+1)\in J$$ If $$f(x) = x^3$$, we have $$1 = 1-x^5 + x^5 = (x^4 + x^3 + x^2+ x + 1)(-x+1) - x^3\cdot x^2 \in J$$ If $$f(x) = x^3+1$$, we have $$1 = -(x^4 + x^3 + x^2+ x + 1)x - (x^3+1)(x^2+x+1)\in J$$ If $$f(x) = x^3+x$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - (x^3+x)(1+x)\in J$$ If $$f(x) = x^3+x+1$$, we have $$1 = (x^4 + x^3 + x^2+ x + 1)(x^2+1) - (x^3+x+1)(x^3+x^2+x)\in J$$ If $$f(x) = x^3+x^2$$, we have $$1 = (x^4 + x^3 + x^2+ x + 1)(-x^2-x+1) - (x^3+x^2)(x^3+x^2+1)\in J$$ If $$f(x) = x^3+x^2+1$$, we have $$1 = (x^4 + x^3 + x^2+ x + 1)(-x^2-x+1) - (x^3+x^2+1)(x^3+x^2)\in J$$ If $$f(x) = x^3+x^2+x$$, we have $$1 = (x^4 + x^3 + x^2+ x + 1)(x^2+1) - (x^3+x^2+x)(x^3+x+1)\in J$$ If $$f(x) = x^3+x^2+x+1$$, we have $$1 = x^4 + x^3 + x^2+ x + 1 - (x^3+x^2+x+1)x\in J$$

• @user26857 Thanks, corrected. The solution now considers all $14$ cases separately. – mechanodroid Feb 18 at 14:51
• So, you want to show that in the polynomial ring $(\mathbb Z/2\mathbb Z)[x]$ the ideal generated by $x^4+x^3+x^2+x+1$ is maximal. But in this case Bezout works since $\mathbb Z/2\mathbb Z$ is a field! – user26857 Feb 18 at 15:01
• @user26857 Yeah, but I wanted my answer to be completely elementary and simple. It seems a bit handwavy to assume the coefficients are in $\mathbb{Z}/2\mathbb{Z}$. And that's basically what the other two answers did. – mechanodroid Feb 18 at 15:04
• If the coefficients are only $0$ and $1$, then you certainly work in the field with two elements. – user26857 Feb 18 at 15:06
• @user26857 But if working in $\mathbb{Z}/2\mathbb{Z}$, while performing euclidean division you may at some point use that $1 = -1$, which isn't true in $\mathbb{Z}$. You would have to formally take the quotient to be able to do that (and I wanted to avoid that route). I wanted to perform the extended euclidean algorithm in $\mathbb{Z}$. – mechanodroid Feb 18 at 15:08