Let $R=\mathbb Z[x]$ be the ring of polynomials over $\mathbb Z$. Prove that the ideal $(x,p)$ is maximal if and only if $p$ is a prime number.

Question

Let $R=\mathbb Z[x]$ be the ring of polynomials over $\mathbb Z$. Prove that the ideal $(x,p)$ generated by $x$ and $p$, where $0<p \in \mathbb Z$, is a maximal ideal of $R$ if and only if $p$ is a prime number. For a prime number $p$, identify the field $R/(x,p)$.

Here is my attempt:

(1) Suppose that $(x,p)$ is maximal.

Assume $p$ is not prime. Then $p = nk$ for some $n, k \in \mathbb Z$.

The elements of $(x,p)$ are of the form $p + p_{1}x + p_{2}x^2 + ...$ where $p_{i}$ are multiples of $p$ for all $i \in \mathbb N$.

$p = kn$ so the elements also have the form $kn + kn_{1}x + kn_{2}x^2 + ...$

But then $(x,p) = (x,kn) \subseteq (x,k)$ and $(x,n)$.

But $(x,p)$ is maximal. Contradiction.

So $p$ is prime.

(2) Now suppose $p$ is prime, and that $(x,p)$ is not maximal.

So there exists a maximal set $M$ such that $(x,p) \subsetneq M \subsetneq R$.

So there exists a polynomial $P \in M$ where $P = n + Q$, $p$ doesn't divide $n$, and $Q$ is some polynomial.

So $(x,p) + P$ is an ideal bigger than $(x,p)$.

But $p$ is prime and doesn't divide $n$, so $gcd(p,n)=1$.

So $1 \in $(x,p) + P$, and is therefore $R$. A contradiction.

So $(X,p)$ must be maximal.

Therefore, $(x,p)$ is a maximal ideal if and only if $p$ is prime.

(3) The field $R/(x,p)$ has elements of the form $(x,p) + P$, where $P \in R, but P \notin (x,p)$. All such $P$ are of the form $n + Q$, $p$ doesn't divide $n$, and $Q$ is some polynomial.

I... think that's it? I'm uncertain about my proofs for (1) and (2), and I'm uncertain what else I need for (3). Any advice would be appreciated!

Answer 1

It is well known that an ideal $I$ in a ring $R$ is maximal iff $R/I$ is a field.

So $(x,p)$ is maximal iff $\mathbb{Z}[x]/(x,p)$ is a field. Also the map $\phi: \mathbb{Z}[x]\rightarrow \mathbb{Z}_p$ defined by $\phi(f(x))=f(0)$ is an epimorphism with the kernal equals to $(x,p)$, so by first isomorphism theorem $\mathbb{Z}[x]/(x,p)\cong \mathbb{Z}_p$. Thus $(x,p)$ is maximal iff $\mathbb{Z}_p$ is a field iff $p$ is a prime.

Answer 2

For (3), the ideal $(x,p)$ contains $x^n$ for each $n\geq 1$, so the only polynomials which survive the quotient are constants. In particular, the field $R/(x,p)$ must be $\mathbb{Z}/p\mathbb{Z}$

Answer 3

Consider the unique ring homomorphism $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}/p\mathbb{Z}$ such that

1. $\varphi(a)=a+p\mathbb{Z}$, for $a\in\mathbb{Z}$,
2. $\varphi(x)=0+p\mathbb{Z}$.

Is $\varphi$ surjective? What's its kernel?

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You can also do that directly, as you attempted. First, how do we characterize the elements of $I=(x,p)$? They are of the form $xf(x)+pg(x)$, so you can prove that $$ f(x)=a_0+a_1x+\dots+a_nx^n\in I \quad\text{if and only if}\quad $a_0\in p\mathbb{Z} $$ (not when every coefficient is a multiple of $p$).

Suppose $p=ab$, with $a>1$ and $b>1$. Then $(x,p)\subsetneq(x,a)$, so $(x,p)$ is not maximal.

If $p$ is prime…