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

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?

• a) You can only work with $a\ne 0$. b) It might happen that $f(0)=1$, might it not? Dec 28, 2015 at 13:52
• What's the contradiction? Dec 28, 2015 at 13:52
• i think i had made a mistake ,$f(0)$ can be equal to 1. Dec 28, 2015 at 13:56
• contraduction is hcf of $(f(0),a)$ is 1 Dec 28, 2015 at 13:58

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 $$\deg\bar f=\deg f\ge1$$.

• The same argument shows that $R[X]$ has no maximal principal ideals whenever $R$ is a UFD having infinitely many non-associate primes. Sep 3, 2016 at 19:09
• One can also generalize this property to $R[X]$, where $R$ is a noetherian integral domain of dimension one having infinitely many maximal ideals; see here. Dec 29, 2019 at 23:38

Main result:

If $$R$$ is an integral domain with infinitely many elements and only finitely many units, then no maximal ideal of $$R[x]$$ is principal.

A pedestrian proof:

Assume $$R$$ is an integral domain with infinitely many elements and only finitely many units.

First, a few basic facts . . .

Since $$R$$ is an integral domain,

• If $$g,h \in R[x]$$ and $$g,h \ne 0$$, then $$\text{deg}(gh) = \text{deg}(g) + \text{deg}(h)\\[4pt]$$.
• If $$r \in R$$, then $$r$$ is a unit in $$R[x]$$ if and only if $$r$$ is a unit in $$R$$.

Also, since $$R$$ is an integral domain, it follows that

• for any $$r \in R$$, and any $$f \in R[x]$$ with $$\text{deg}(f) \ge 1$$, the equation $$f(x) = r$$ has only finitely many roots in $$R$$.

Next, some lemmas . . .

Lemma $$\mathbf{1}$$:

If $$a,b \in R$$ and $$a$$ is not a unit in $$R$$, then $$(a,x-b)$$ is a proper ideal of $$R[x]$$.

proof:

Suppose instead that $$(a,x-b) = (1)$$.

\begin{align*} \text{Then}\;\,&(a,x-b) = (1)\\[4pt] \implies\; &ag(x) + (x-b)h(x) = 1,\;\text{for some}\;g,h \in R[x]\\[4pt] \implies\; &ag(b) + (b-b)h(b) = 1,\;\text{for some}\;g,h \in R[x]\\[4pt] \implies\; &ag(b) = 1,\;\text{for some}\;g \in R[x]\\[4pt] \implies\; &a\;\text{is a unit in R}\\[4pt] \end{align*}

This completes the proof of lemma $$1$$.

Lemma $$\mathbf{2}$$:

If $$a \in R$$, the ideal $$(a)$$ of $$R[x]$$ is not a maximal ideal.

proof:

Suppose instead that for some $$a \in R$$, the ideal $$(a)$$ of $$R[x]$$ is a maximal ideal of $$R[x]$$.

Since $$(a)$$ is maximal in $$R[x]$$, $$(a) \ne (1)$$, hence $$a$$ is not a unit of $$R$$.

Since $$a$$ is not a unit of $$R$$, it follows that $$x \notin (a)$$.

Since $$(a)$$ is maximal, and $$x \notin (a)$$, it follows that $$(a,x) = (1)$$, which contradicts lemma $$1$$, since $$a$$ is not a unit of $$R$$.

This completes the proof of lemma $$2$$.

proof of the main result:

Suppose the principal ideal $$(f) \in R[x]$$ is maximal, for some $$f \in R[x]$$.

Our goal is to derive a contradiction.

By lemma $$2$$, $$f$$ has degree at least $$1$$, hence $$(f)$$ has no nonzero constants elements.

Since $$R$$ has infinitely many elements but only finitely many units, there exists an element $$b \in R$$, such that $$f(b)$$ is a nonzero nonunit. Actually, there are infinitely many such elements $$b$$, but we only need one.

Thus, suppose $$b \in R$$ is such that $$f(b) = a$$, where $$a \in R$$ is a nonzero nonunit.

\begin{align*} \text{Then}\;\, &\text{deg}(f) \ge 1\\[4pt] \implies\; &f(x) = f(b) + (x-b)g(x),\;\text{for some nonzero }g \in R[x]\\[4pt] \implies\; &(f,a) \subseteq (a,x-b)\\[4pt] \implies\; &(f,a) \ne (1)\qquad\text{[since by lemma 1, (a,x-b) \ne (1)]}\\[4pt] \implies\; &(f,a) = (f)\qquad\text{[since (f) is maximal]}\\[4pt] \implies\; &a \in (f)\\[4pt] \end{align*}

contradiction, since $$(f)$$ has no nonzero constants elements.

This completes the proof of the main result.

Corollary:

No maximal ideal of $$\mathbb{Z}[x]$$ is principal.

proof:

This follows from the main result since $$\mathbb{Z}$$ is an infinite integral domain with only two units, namely $$\pm 1$$.

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 $\langle f(x)\rangle\subsetneq\langle f(x),p\rangle\subsetneq\Bbb Z[x]$.

• Are you showing $f(x)g(x)+ph(x)\not=1$, where $1$ is the identity, and proving there are no units? Or that $f(x)g(x)+ph(x)\not=1$ and $1\in\Bbb Z[x]$ thus $\langle f(x),p\rangle\subsetneq\Bbb Z[x].$ I am also still confused why $p$ can't divide that leading coefficient of $f$, what would happen for example if $f(x)=14x$ and $p=7?$ Oct 9, 2016 at 22:31
• $p$ was chosen to be a prime not dividing the lead coefficient of $f$. This is in part to force $p$ to divide the leading coefficient of $g(x)$ in the expression $f(x)g(x)+ph(x)$ Nov 18, 2022 at 2:25

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 $p$.