Prime ideal

An ideal $\mathfrak{p}$ is prime when $ab\in \mathfrak{p}$ implies $a\in \mathfrak{p}$ or $b\in \mathfrak{p}$, equivalently when the quotient is an integral domain.

A minimal prime ideal is a prime ideal containing no strictly smaller prime ideal. If $A$ is a reduced ring and $\mathfrak{p}$ is minimal, the localization at a prime ideal $A_{\mathfrak{p}}$ is a field: its only prime is its maximal ideal, and reducedness makes that ideal its zero nilradical. Every nonzero commutative ring has a minimal prime.

Every proper ideal of a unital commutative ring is contained in a minimal prime over that ideal. A maximal ideal supplies a prime containing it, and an intersection of a decreasing chain of primes is still prime. Zorn's lemma, with reverse inclusion, therefore supplies a minimal member. No Noetherian assumption is needed.

In $k[u,v]$, the ideal $I=(u^2,uv)$ has radical of an ideal $(u)$, a prime ideal, and hence a unique minimal prime ideal. Nevertheless $uv\in I$, $u\notin I$, and $v^j\notin I$ for all $j\ge 1$, so $I$ is not a primary ideal. The same example works after adjoining inverses of $1+u$ and $1+v$; thus it also gives an example in a two-variable Laurent polynomial ring.

In a Noetherian ring, an ideal $I$ disjoint from a nonempty multiplicative subset $S$ extends to an ideal $P$ maximal among those disjoint from $S$, by the ascending chain condition. It is proper. If $ab\in P$ but neither factor belongs to $P$, both $P+(a)$ and $P+(b)$ meet $S$. Multiplying such representatives puts an element of $S$ in $P$, a contradiction. Thus $P$ is a prime ideal. For a general commutative ring the same argument follows after applying the Zorn lemma to disjoint ideals; the union of a chain is disjoint from $S$.

The $n$th symbolic power of a prime ideal $P\subset R$ iswhere the intersection denotes contraction of an ideal. It is always a $P$-primary ideal. The ordinary power $P^n$ equals $P^{(n)}$ precisely when $P^n$ is $P$-primary; this assertion does not require $R$ to be Noetherian.

For a commutative unital ring $R$, an ideal $I$ is prime exactly when $R/I$ is an integral domain. The zero-product condition in the quotient translates directly to $ab\in I$.