Nonconstant polynomials do not generate maximal ideals in $\mathbb Z[x]$

Question

Let $f$ be a nonconstant element of ring $\mathbb Z[x]$. Prove that $\langle f \rangle$ is not maximal in $\mathbb Z[x]$.

Let us assume $\langle f \rangle$ is maximal. Then $\mathbb Z[x] / \langle f \rangle$ would be a field. Let $a \in \mathbb{Z}$. Then $a + \langle f \rangle$ is a nonzero element of this field, hence a unit. Let $g + \langle f \rangle$ be its inverse. Then $a g - 1 \in \langle f \rangle$, hence $ag(x)-1 = f(x)h(x)$ for some $h \in Z[x]$, hence $ag(0) + f(0)h(0) = 1$, thus $(a,f(0))=1$ for all $a \in \Bbb Z$, contradiction, hence the proof.

Is my argument correct? Is there any other method?

Answer 1

Let $p\in\mathbb Z$ be a prime such that $p\nmid\text{LC}(f)$, where $\text{LC}(f)$ stands for the leading coefficient of $f$. Moreover $p$ is non-zero in $\mathbb Z[x]/(f)$, hence invertible in $\mathbb Z[x]/(f)$, so there are $g,h\in\mathbb Z[x]$ such that $pg(x)+f(x)h(x)=1$. It follows that $\bar f\bar h=\bar 1$ in $(\mathbb Z/p\mathbb Z)[x]$, and this is impossible since $\deg\bar f=\deg f\ge1$.

Answer 2

Let $f(x)\in\Bbb Z[x]$ have degree greater than zero. Choose a prime $p$ that does not divide the leading coefficient of $f$. Then $p\not=f(x)g(x)$ for any $g(x)\in\Bbb Z[x]$ (because $\deg(fg)=\deg(f)+\deg(g)$) and $f(x)g(x)+ph(x)\not=1$ for all $g(x),h(x)\in\Bbb Z[x]$ (consider the coefficients of $g$ in descending order starting with the leading coefficient to see that $p$ would have to divide every coefficient of $g$ and therefore would have to divide one). Thus $\langle f(x)\rangle\subsetneq\langle f(x),p\rangle\subsetneq\Bbb Z[x]$.

Answer 3

Claim: $ \frac{{\bf Z}[x]}{(f(x))}$ is not a field.

Proof: Let $a \in \bf Z$ be such that $f(a)$ is not equal to $0, ±1$ and choose a prime $p$ dividing $f(a)$. Consider $ \pi : {\bf Z}[x] \to\frac {\bf Z}{(p)}$ be the unique homomorphism with $\pi (x) = a \bmod p$. Then $\pi$ factors through $ \frac {{\bf Z}[x]}{(f(x))}$ since $ \pi (f(x)) = 0.$ Now, $ \frac {{\bf Z}[x]}{(f(x))}$ is infinite, so $\pi: \frac {{\bf Z}[x]}{(f(x))} \to \frac {{\bf Z}}{(p)}$ is not injective. If we show that $\pi$ is not the zero map, then $\ker\pi$ will be a non-trivial ideal of $ \frac {{\bf Z}[x]}{(f(x))}$ and it won’t be a field. If $\pi$ is the zero map, then $\pi(1) = 0$, i.e., there exists polynomials $u, v ∈ {\bf Z}[x] $ with $1 = u(x)f(x) + pv(x)$. Putting $x = a$ we get a contradiction since $u(a)f(a) + pv(a)$ is divisible by $p$ as well as being equal to $1$.

Moreover, it can be proved that maximal ideals of ${\bf Z}[x]$ are precisely of the form $(p,f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in ${\bf Z}[x]$ which is irreducible modulo $p$.