Every prime ideal of the integers is a maximal ideal. Assume the ideal is not $⟨0⟩$.

Question

The book is giving the hint: If $⟨p⟩ ⊆ ⟨a⟩$, but $⟨p⟩ ≠ ⟨a⟩$, explain why $gcd(p, a) = 1$ and conclude that $1 ∈ ⟨a⟩$. However, I cannot seem to get anywhere with this. I know the definition of both prime and maximal ideals. I also know that $⟨p⟩$ is a prime ideal iff $p$ is a prime number. Thank you!

Answer 1

If $\langle a\rangle\not\subseteq\langle p\rangle$, then there exists $b$ such that $ab\not\in\langle p\rangle$, i.e. $p$ does not divide $ab$. As a result, $p$ does not divide $a$, so $\gcd(p,a)=1$.

Now $\gcd(p,a)$ can be written as a linear combination $ax+py$ of $a$ and $p$, so $1=\gcd(p,a)\in\langle a\rangle$.

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Hope this helps.

Answer 2

If $(p) \subseteq (a)$, then $a|p$, and so $a=1$ or $a=p$. but $(a) \ne (p),$ then $(a)=(1)$