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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$1 = x^4 + x^3 + x^2+ x + 1 - (x^3+x^2+x+1)x\in J$$