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.

The prime ideals $P$ of $\mathbb{Z}[t]$ with $P\cap \mathbb{Z}=\{0\}$ are the zero ideal and the principal ideals $(f)$ generated by nonconstant primitive polynomials that are irreducible in $\mathbb{Q}[t]$.

To prove the description, invert the nonzero integers to obtain the localization $\mathbb{Q}[t]$. The extension of an ideal $P$ consists of the polynomials $f/n$ with $f\in P$ and $n>0$. Its contraction is $P$: if $nh\in P$ and $n\ne 0$, the defining condition puts $n\notin P$, so primeness gives $h\in P$. The prime ideal correspondence for localization makes this extension a proper prime ideal of $\mathbb{Q}[t]$.

If $P\ne 0$, its extension has a nonzero generator because $\mathbb{Q}[t]$ is a principal ideal domain. Dividing this generator by its rational polynomial content gives an integer-coefficient primitive polynomial $f$ generating the same extended ideal. Divisibility descent by Gauss lemma for polynomials then gives $P=(f)$ in $\mathbb{Z}[t]$. The proper prime ideal $(f)$ in $\mathbb{Q}[t]$ makes $f$ irreducible there. Conversely, any such polynomial generates a prime ideal in $\mathbb{Q}[t]$ whose contraction is $(f)\subset \mathbb{Z}[t]$ and whose intersection with $\mathbb{Z}$ is zero.

The intersection condition is essential: $(2,t)$ is a maximal ideal of $\mathbb{Z}[t]$, since its quotient ring is ${\mathbb{F}}_2$, but it is not principal. A common divisor of $2$ and $t$ must be a unit, which would generate the whole ring.

Let $R$ be a unique factorization domain and let $p$ be a prime element. The only prime ideals contained in the principal ideal $(p)$ are $(0)$ and $(p)$. To prove this, suppose $0\ne P⊊(p)$. Then $p\notin P$. For $0\ne f\in P$, unique factorization allows us to write $f=p^{k}g$ with $g\notin (p)$. Since $P$ is a prime ideal and $p\notin P$, repeated primality gives $g\in P\subseteq (p)$, a contradiction.

In the particular polynomial ring $\mathbb{Z}[t]$, when $p$ is a prime number, the same proof uses only the integer coefficients: factor out the largest common power of $p$ from the coefficients of a nonzero polynomial $f$. This gives $f=p^{k}g$ with $g\notin (p)$ directly, without requiring the general unique factorization domain theorem for polynomial rings.

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.

A prime ideal $P$ is minimal over an ideal $I$ if $I\subseteq P$ and no smaller prime ideal contains $I$. The definition also applies to two-sided prime ideals of a noncommutative ring.

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 $ab\in I$.