$p\in \mathbb Z$ is prime if and only if $\langle p\rangle$ is maximal. [duplicate]

$p\in \mathbb Z$ is prime if and only if $\langle p\rangle$ is maximal.

Could someone give me a hint to prove it. please?

• math.stackexchange.com/a/772115/227902 – Decaf-Math Jul 14 '17 at 2:04
• A more general approach would be to prove that an ideal $I$ is maximal $\iff$ the quotient $R/I$ is a field. This is an important theorem, and can be applied here; indeed $\mathbb{Z}/p \mathbb{Z}$ is a field. – Kaj Hansen Jul 14 '17 at 2:07
• @KajHansen yes but at this moment I'm just interested about the particular case :) – user441848 Jul 14 '17 at 2:12
• @AnneliseToft what have you tried? – vociferous_rutabaga Jul 14 '17 at 2:12
• @Annelise Toft: As an experienced MSE participant, you are expected to make an effort to state the question unambigously. In your initial post, no ring was mentioned, so a reader would have had to guess your intention. As you can see, an edit was needed, and was supplied by Pedro Tamaroff. – quasi Jul 14 '17 at 2:14