Let $R = \mathbb{Z}[x]$ and $n\geq 1$.
Show that $(n,x)$ is a prime ideal if and only if $n$ is a prime number.
My proof:
$$\mathbb{Z}[x] = \{ a_o + a_1x^1 + ... + a_nx^n: n \in \mathbb{N}_0, a_0,...,a_n \in \mathbb{Z} \}$$
Since $\mathbb{Z}[x]$ is a commutative ring,
$$(n,x) = (n) + (x)$$
where
$$(n) = \{ nk: k \in \mathbb{Z}\}, (x) = \{ a_1x + a_2x^2 + ... + a_nx^2: n\in \mathbb{N}, a_1,...,a_n \in \mathbb{Z}\}$$
Hence,
$$(n) + (x) = \{ nk + a_1x + a_2x^2 + ... + a_nx^n: n\in \mathbb{N}, a_1,...,a_n \in \mathbb{Z}\}$$
Let $p, q \in \mathbb{Z}[x]$ and $pq \in (n,x)$. Then we can write $p,q$ and $pq$ as $$ p = p_0 + p_1x + ... + p_mx^m, \mbox{ for some } m \in \mathbb{N} $$ $$ q =q_0 + q_1x + ... + q_rx^r \mbox{ for some } r \in \mathbb{N} $$ and their product as $$ pq = p_0q_0 + c_1x + ... + c_kx^k, \mbox{ for } k = m + r, c_1,...,c_k \in \mathbb{Z}$$
By hypothesis $(n,x)$ is a prime ideal and $pq \in (n,x)$ hence
$$p \in (n,x), \mbox{ or }, q \in (n,x)$$
If $p \in (n,x)$ then $$p = nk_p + p_1x + p_2x^2 + ... + p_mx^m, \mbox{ for some } k_p \in \mathbb{Z}$$ Hence, $nk_p = p_0 \Rightarrow n|p_0$ or $n=1$ and $k_p = p_0$. If $n = 1$ we have that $(n,x) = \mathbb{Z}[x]$. If $(n,x) = \mathbb{Z}[x]$ we contradict the assumption that $(n,x)$ is a prime ideal because prime ideals are proper. Hence, $n$ can not be one. We pick $p_0$ as an arbitrary integer, and $n$ divides $p_0$, hence $n$ must be prime.
It remains to proof that $n$ is prime if $q \in (n,x)$, but the proof is the same if we replace $p$ by $q$.
Now I dont know how to prove the other direction.
Can anyone verify this proof for me, and help me with the other direction?
Thanks!
Hint: Show that $$\Bbb Z[x]/(n,x)\simeq \Bbb Z/n\Bbb Z$$ This gives you the equivalence right away, because an ideal is prime if and only if the quotient is...