Prime ideal

Last updated January 27, 2022

Definition

Let $A$ be a (commutative) ring.

Definition. An ideal $p\subset A$ is called prime if the following equivalent conditions are met:

1. For every $a,b\in A$ such that $ab\in p$ either $a\in p$ or $b\in p$.
2. For every $a\not\in p$ and $b\not\in p$ then $ab\not\in p$.
3. $A/p$ is an integral domain.

Theorem. The intersection of all prime ideals of $A$ is the nilradical $Nil(A)$.

Over / under

Integral extensions

( integral extensions)

Let $\mathcal{O}$ be a dedekind domain, $k$ its field of fractions and $K/k$ a finite separable extension. Let $\mathcal{O}_K$ be the integral closure.

The prime ideal $\mathfrak{B}\subset\mathcal{O}_K$ lies over the prime ideal $\mathfrak{p}\subset\mathcal{O}$, denoted as $\mathfrak{B}\mid\mathfrak{p}$, iff $$\mathfrak{B}\cap\mathcal{O}=\mathfrak{p}.$$

( extensions of p-adic absolute values)

Properties

1. The zero ideal is prime if and only if $A$ is an integral domain.
2. If $f:A\to B$ is a ring homomorphism, and $\mathfrak{q}$ is a prime ideal in $B$, then $f^{-1}(\mathfrak{q})$ is a prime ideal in $A$.
3. TFAE:
   1. $\mathfrak{q}$ lies over $\mathfrak{p}$
   2. $\mathfrak{p}\subset\mathfrak{q}$, by which we mean $\mathfrak{p}S\subset\mathfrak{q}$
   3. $\mathfrak{q}$ occurs in the factorization of $\mathfrak{p}S$.

Proofs

1

For the forward direction, suppose the zero ideal is prime. Then if $xy=0$ either $x=0$ or $y=0$. This by definition makes $A$ an integral domain. Convsersely, suppose $A$ is an integral domain. Then if $xy=0$ either $x=0$ or $y=0$. This makes $(0)$ a prime ideal.

2

$f^{-1}(\mathfrak{q})$ is the kernel of the quotient map $A\to B/f^{-1}(\mathfrak{q})$, hence $A/f^{-1}(\mathfrak{q})\cong B/\mathfrak{q}$. The latter is an integral domain since $\mathfrak{q}$ is prime, so the former is as well, meaning $f^{-1}(\mathfrak{q})$ is a prime ideal.