Short Question: $(p)$ for a prime is not a maximal ideal in $\mathbb{Z}[X]$

Question

Given a prime number $p\in\mathbb{Z}$ I want to show that $(p)$ is not a maximal ideal in the ring of polynomials $\mathbb{Z}[X]$.

I know how the maximal ideals in $\mathbb{Z}[X]$ look like, I want to explicitly show that $(p)$ is not maximal.

Answer 1

What does the ideal

$(p) \subset \Bbb Z[x] \tag 1$

"look like"? I claim that

$(p) = \left \{ f(x) = \displaystyle \sum_0^{\deg f} f_ix^i \in \Bbb Z[x], \; p \mid f_i, 0 \le i \le n \right \}; \tag 2$

that is, $(p)$ consists of those $f(x) \in \Bbb Z[x]$ whose coefficients are all multiples of $p \in \Bbb P$; this is nearly self-evident since the members of $(p)$ are all of the form $pg(x)$ where $g(x) \in \Bbb Z[x]$;to see that $(p)$ is not maximal, we consider the ideal

$J = \left \{ g(x) = \displaystyle \sum_0^{\deg g} g_ix^i \in \Bbb Z[x], \; p \mid g_0 \right \} \subset \Bbb Z[x]; \tag 3$

the elements of $J$ are those $g(x) \in \Bbb Z[x]$ whose constant term is a multiple of $p$, with no constraint on the $g_i$, $1 \le i \le n$.

It is easy to see that both $(p)$ and $J$ are ideals in $\Bbb Z[x]$; indeed, for

$h(x) = \displaystyle \sum_0^n h_i x^i, k(x) = \sum_0^n k_ix^i \in J \tag 4$

we have

$p \mid h_0, p \mid k_0 \Longrightarrow p \mid (h_0 - k_0) \tag 5$

which shows that

$h(x) - k(x) = \displaystyle \sum_0^n (h_i - k_i)x^i \in J; \tag 6$

also, with

$h(x) \in J, k(x) \in \Bbb Z[x] \tag 7$

we have

$(h(x)k(x))_0 = h_0k_0, \tag 8$

and

$p \mid h_0 \Longrightarrow p \mid h_0k_0 \Longrightarrow h(x)k(x) \in J, \tag 9$

showing that $J$ is an ideal in $\Bbb Z[x]$; verification that $(p)$ is also an ideal in $\Bbb Z[x]$ is quite similar to this and will be left to the reader.

We observe that $J$ is a proper ideal; that is,

$J \subsetneq \Bbb Z[x]; \tag{10}$

this is easily seen by virtue of the fact that the constant term of a general polynomial in $\Bbb Z[x]$ may take any integer value whatsoever, whereas those of members of $J$ must be divisible by $p$; and again, a similar argument shows that

$(p) \subsetneq J, \tag{11}$

since $p$ divides the coefficients of the non-constant terms of the elements of $(p)$, whereas those of members of $J$ may be arbitary elements of $\Bbb Z$. Therefore we find that

$(p) \subsetneq J \subsetneq \Bbb Z[x], \tag{12}$

so $(p)$ cannont be maximal.

Note Added in Edit, Friday 21 June 2019 10:16 PM PST: We in fact have

$J = (p, x); \tag{13}$

the proof is quite simple. From here, it is easy to see that (12) is essentially the same as the proper inclusion of ideals given by Ruben du Burck in his answer. End of Note.

Answer 2

We have the proper inclusion $(p) \subset (p,X)$.