Ring Theory 02: Prime Ideals, Maximal Ideals, and Localization

Ring theory / 02

Prime ideals, maximal ideals, and localization

Maximal ideals recover ordinary closed points in familiar polynomial rings, but prime ideals are stable under pullback and carry the generic information needed for geometry.

A crucial early question is why scheme theory uses prime ideals rather than only maximal ideals. Atiyah and Macdonald give the algebraic answer: prime ideals behave well under inverse image, localization, radicals, and local properties.

Prime and maximal ideals

An ideal \(\mathfrak p\subset A\) is prime if \(\mathfrak p\ne A\) and

\[ab\in\mathfrak p\Rightarrow a\in\mathfrak p\text{ or }b\in\mathfrak p.\]

An ideal \(\mathfrak m\subset A\) is maximal if \(\mathfrak m\ne A\) and no ideal lies strictly between \(\mathfrak m\) and \(A\).

Algebraic tests

For an ideal \(I\subset A\), with integral domains and fields understood to be nonzero:

\[I\text{ is prime}\quad\Longleftrightarrow\quad A/I\text{ is an integral domain},\]

and

\(I\text{ is maximal}\quad\Longleftrightarrow\quad A/I\text{ is a field}.\)

The second equivalence implies every maximal ideal is prime. The converse is false: \((0)\subset k[x]\) is prime because \(k[x]\) is a domain, but it is not maximal because \(k[x]\) is not a field.

Proofs of the algebraic tests

For primality, assume first that \(I\) is prime. If

\[(a+I)(b+I)=0\]

in \(A/I\), then \(ab\in I\). Since \(I\) is prime, \(a\in I\) or \(b\in I\), so \(a+I=0\) or \(b+I=0\). Thus \(A/I\) is an integral domain. Conversely, if \(A/I\) is a domain and \(ab\in I\), then

\[(a+I)(b+I)=0\]

in \(A/I\). Since a domain has no zero divisors, \(a+I=0\) or \(b+I=0\), so \(a\in I\) or \(b\in I\). Hence \(I\) is prime.

For maximality, assume \(I\) is maximal and \(a+I\ne0\). Then \(a\notin I\), so the ideal \(I+(a)\) strictly contains \(I\) and therefore equals \(A\). Thus \(i+ba=1\) for some \(i\in I\), \(b\in A\), so \(b+I\) is the inverse of \(a+I\). Hence \(A/I\) is a field. Conversely, if \(A/I\) is a field and \(J\) is an ideal with \(I\subseteq J\subseteq A\), then either \(J=I\), or \(J\) contains some \(a\notin I\). In the latter case \(a+I\) is invertible, so \(1\in J\), and \(J=A\). Hence \(I\) is maximal.

Maximal ideals and classical points

For an algebraically closed field \(k\) and finite \(n\), the weak Nullstellensatz says every maximal ideal of \(k[x_1,\ldots,x_n]\) has the form

\[(x_1-a_1,\ldots,x_n-a_n)\]

for a point \((a_1,\ldots,a_n)\in k^n\). This is the classical reason maximal ideals look like \(k\)-rational closed points in affine space.

But prime ideals include more information. In \(k[x]\),

\[\operatorname{Spec}k[x]=\{(0)\}\cup\{(x-a):a\in k\}\]

when \(k\) is algebraically closed. The point \((0)\) is the generic point of the whole affine line: its closure is the entire space.

Example: real polynomial points

In \(\mathbb R[x]\), maximal ideals include \((x-a)\) for \(a\in\mathbb R\), but also ideals generated by irreducible quadratic polynomials such as \((x^2+1)\). The quotient \(\mathbb R[x]/(x^2+1)\cong\mathbb C\) is a field. Thus even maximal ideals need not be literal real-valued points.

Residue values on maximal spectra

Write \(\operatorname{mSpec}A\) for the set of maximal ideals of \(A\). Maximal ideals recover the most familiar form of point evaluation. If \(\mathfrak m\) is maximal, then \(A/\mathfrak m\) is a field, called the residue field at the closed point \(\mathfrak m\). An element \(a\in A\) gives a set-theoretic residue-value map

\[a:\operatorname{mSpec}A \to \coprod_{\mathfrak m\in\operatorname{mSpec}A}A/\mathfrak m, \qquad \mathfrak m\mapsto a\bmod\mathfrak m.\]

The target is a disjoint union because different closed points may have different residue fields; the value at \(\mathfrak m\) lies in the \(A/\mathfrak m\)-component of that disjoint union.

For \(A=\mathbb C[x]\), the maximal ideals are \((x-\alpha)\), and an element \(f\in\mathbb C[x]\) gives

\[f:\operatorname{mSpec}\mathbb C[x] \to \coprod_{\alpha\in\mathbb C}\mathbb C[x]/(x-\alpha), \qquad (x-\alpha)\mapsto f\bmod(x-\alpha)=f(\alpha).\]

Each quotient \(\mathbb C[x]/(x-\alpha)\) is canonically \(\mathbb C\), so this is exactly the ordinary polynomial function \(\alpha\mapsto f(\alpha)\).

For \(A=\mathbb Z\), the maximal ideals are \((p)\), where \(p\) is prime. An integer \(n\in\mathbb Z\) gives

\[n:\operatorname{mSpec}\mathbb Z \to \coprod_{p\text{ prime}}\mathbb Z/(p), \qquad (p)\mapsto n\bmod p.\]

Thus an integer has one residue value modulo every prime. This is already less like ordinary real- or complex-valued calculus: the codomain varies with the point.

From maximal spectra to full spectra

For an arbitrary prime \(\mathfrak p\subset A\), the quotient \(A/\mathfrak p\) is a domain, not necessarily a field. Its fraction field

\[\kappa(\mathfrak p)=\operatorname{Frac}(A/\mathfrak p)\]

is the residue field at \(\mathfrak p\). Every element \(a\in A\) has a value

\[a(\mathfrak p)\in\kappa(\mathfrak p), \qquad a(\mathfrak p)=a\bmod\mathfrak p.\]

For \(\operatorname{Spec}\mathbb C[x]\), the closed point \((x-\alpha)\) has residue field \(\mathbb C\), while the generic point \((0)\) has residue field \(\mathbb C(x)\). At the generic point, a polynomial is not evaluated at a complex number; it is regarded as an element of the rational function field.

For \(\operatorname{Spec}\mathbb Z\), the closed point \((p)\) has residue field \(\mathbb F_p\), while the generic point \((0)\) has residue field \(\mathbb Q\). An integer therefore has both all of its reductions modulo primes and its generic rational value.

This is one reason \(\operatorname{Spec}A\), not only \(\operatorname{mSpec}A\), is the correct space. The maximal spectrum records closed-point values. The full spectrum also records generic behavior, irreducible components, specialization, localization, and functoriality under ring maps.

Why primes are functorial

Let \(\varphi:A\to B\) be a unital ring homomorphism and let \(\mathfrak q\subset B\) be prime. Then

\[\varphi^{-1}(\mathfrak q)\subset A\]

is prime.

Proof

The inverse image is proper because \(1_B\notin\mathfrak q\). If \(ab\in\varphi^{-1}(\mathfrak q)\), then

\[\varphi(a)\varphi(b)=\varphi(ab)\in\mathfrak q.\]

Since \(\mathfrak q\) is prime, \(\varphi(a)\in\mathfrak q\) or \(\varphi(b)\in\mathfrak q\). Therefore \(a\in\varphi^{-1}(\mathfrak q)\) or \(b\in\varphi^{-1}(\mathfrak q)\).

Maximal ideals do not have this stability. For example, the inclusion \(\mathbb Z\hookrightarrow\mathbb Q\) pulls back the maximal ideal \((0)\subset\mathbb Q\) to \((0)\subset\mathbb Z\), which is prime but not maximal.

This is the categorical reason scheme points are prime ideals: maps of rings must produce continuous maps of spectra.

Multiplicative sets and localization

A subset \(S\subset A\) is multiplicative if \(1\in S\) and \(s,t\in S\Rightarrow st\in S\). The localization \(S^{-1}A\) is the ring whose elements are fractions

\[\frac a s,\qquad a\in A,\ s\in S,\]

where elements of \(S\) become invertible.

The universal property says: a homomorphism \(u:A\to B\) factors uniquely as

\[A\to S^{-1}A\to B\]

precisely when every element of \(S\) maps to a unit of \(B\).

Important cases:

\[A_f=A[\frac1f]\quad\text{where }S=\{1,f,f^2,\ldots\},\]

and

\[A_{\mathfrak p}=(A\setminus\mathfrak p)^{-1}A.\]

Local rings

A ring \(R\) is local if it has a unique maximal ideal. For a prime \(\mathfrak p\), the localization \(A_{\mathfrak p}\) is local, with maximal ideal

\[\mathfrak pA_{\mathfrak p} = \left\{\frac a s:a\in\mathfrak p,\ s\notin\mathfrak p\right\}.\]

Proof

Every element of \(A_{\mathfrak p}\) has the form \(a/s\) with \(s\notin\mathfrak p\). If \(a\notin\mathfrak p\), then \(a\) is also an allowed denominator, so \(a/s\) has inverse \(s/a\). If \(a\in\mathfrak p\), then \(a/s\in\mathfrak pA_{\mathfrak p}\). Thus every element is either a unit or lies in \(\mathfrak pA_{\mathfrak p}\).

The set \(\mathfrak pA_{\mathfrak p}\) is an ideal, and the quotient is

\[A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}\cong\operatorname{Frac}(A/\mathfrak p),\]

because localization inverts exactly the nonzero elements of the domain \(A/\mathfrak p\). Hence \(\mathfrak pA_{\mathfrak p}\) is maximal. Any maximal ideal contains no units, so by the first paragraph it must be contained in \(\mathfrak pA_{\mathfrak p}\); maximality then forces equality. Therefore \(A_{\mathfrak p}\) has a unique maximal ideal.

Geometrically, \(A_{\mathfrak p}\) is the ring of functions defined near the point \(\mathfrak p\). It remembers only behavior in neighborhoods of that point.

Support of a module

For an \(A\)-module \(M\), the support is

\[\operatorname{Supp}M=\{\mathfrak p\in\operatorname{Spec}A:M_{\mathfrak p}\ne0\}.\]

If \(M=A/I\), then

\[\operatorname{Supp}(A/I)=V(I).\]

Indeed, \((A/I)_{\mathfrak p}\cong A_{\mathfrak p}/IA_{\mathfrak p}\). This localized quotient is zero exactly when \(IA_{\mathfrak p}=A_{\mathfrak p}\). If \(I\not\subseteq\mathfrak p\), choose \(i\in I\setminus\mathfrak p\); then \(i\) is a unit in \(A_{\mathfrak p}\), so \(IA_{\mathfrak p}=A_{\mathfrak p}\). Conversely, if \(IA_{\mathfrak p}=A_{\mathfrak p}\), then

\[1=\sum_{j=1}^n\frac{i_j}{s_j}\]

with \(i_j\in I\) and \(s_j\notin\mathfrak p\). Multiplying by \(s=s_1\cdots s_n\notin\mathfrak p\) gives

\[s/1= \left(\sum_{j=1}^n i_j\prod_{\ell\ne j}s_\ell\right)/1\]

in \(A_{\mathfrak p}\). Therefore there exists \(t\notin\mathfrak p\) such that

\[t\left(s-\sum_{j=1}^n i_j\prod_{\ell\ne j}s_\ell\right)=0\]

in \(A\). Hence \(ts\in I\). Since both \(t\) and \(s\) are outside the prime ideal \(\mathfrak p\), their product \(ts\) is outside \(\mathfrak p\). Thus \(I\) contains an element not in \(\mathfrak p\), so \(I\not\subseteq\mathfrak p\). Hence \((A/I)_{\mathfrak p}\ne0\) exactly when \(I\subseteq\mathfrak p\), and the module \(A/I\) lives exactly on the closed set cut out by \(I\). This is the module-theoretic predecessor of coherent sheaves supported on closed subschemes.

Common confusion

Localization is not the same as quotienting. Quotienting by \(I\) imposes equations and moves to a closed subset. Localizing at \(f\) inverts \(f\) and moves to the open subset where \(f\) does not vanish.