Search the corpus

    Commutative algebra

    Contents

    Throughout, rings are commutative with identity.

    Zorn’s lemma

    Chain in a poset

    Let \((X,\leq)\) be a partially ordered set. A chain in \(X\) is a subset \(C\subseteq X\) that is totally ordered by \(\leq\): for all \(c,c'\in C\), \(c\leq c'\) or \(c'\leq c\).

    Let \((X,\leq)\) be a partially ordered set and \(S\subseteq X\). An upper bound of \(S\) is an element \(x\in X\) such that \(s\leq x\) for all \(s\in S\).

    Zorn's lemma

    Upper bounds for every chain force a maximal element.Suggest a slogan

    If \(P\) is a poset in which every chain has an upper bound in \(P\), then \(P\) has a maximal element.

    Zorn's lemma

    If every chain has an upper bound, a maximal element exists.Suggest a slogan

    Let \((P,\le)\) be a partially ordered set in which every chain has an upper bound in \(P\). Then \(P\) has a maximal element: an element \(m\in P\) such that \(m\le x\) with \(x\in P\) implies \(x=m\).

    The empty chain is a chain, so the hypothesis forces \(P\) to be nonempty.

    State Zorn's lemma.

    Krull's theorem on maximal ideals

    Every proper ideal of a nonzero ring sits inside a maximal ideal.Suggest a slogan

    Let \(R\) be a nonzero ring. Then every proper ideal of \(R\) is contained in a proper maximal ideal; in particular, \(R\) has a proper maximal ideal.

    Krull's theorem on maximal ideals

    Every proper ideal extends to a maximal ideal.Suggest a slogan

    Let \(R\) be a commutative ring. Every proper ideal \(I\subsetneq R\) is contained in a maximal ideal of \(R\). In particular, applying this to \(I=0\), every nonzero commutative ring has a maximal ideal.

    Let \(P\) be the set of proper ideals of \(R\) containing \(I\), ordered by inclusion; it contains \(I\). The union of a nonempty chain in \(P\) is an ideal containing \(I\), and it is proper because it does not contain \(1\), so it is an upper bound in \(P\); the empty chain has the upper bound \(I\). By Zorn's lemma, \(P\) has a maximal element, which is a maximal ideal of \(R\) containing \(I\).

    What is Krull's theorem?

    Existence of maximal ideals

    Every proper ideal is contained in a maximal ideal.Suggest a slogan

    Let \(R\) be a ring and \(I\subsetneq R\) a proper ideal. Then \(I\) is contained in a maximal ideal of \(R\).

    Let \(\mathcal S\) be the set of proper ideals of \(R\) containing \(I\), ordered by inclusion; it is nonempty since \(I\in\mathcal S\). If \(\mathcal C\subseteq\mathcal S\) is a nonempty chain, then \(J\coloneqq\bigcup_{J'\in\mathcal C}J'\) is an ideal containing \(I\), and \(1\notin J\) because \(1\) lies in no proper ideal, so \(J\in\mathcal S\) is an upper bound for \(\mathcal C\). By Zorn’s lemma \(\mathcal S\) has a maximal element \(\mathfrak m\). Any ideal strictly containing \(\mathfrak m\) contains \(I\) and is not in \(\mathcal S\), so it equals \(R\); thus \(\mathfrak m\) is a maximal ideal.

    Every non-unit of $R$ is contained in a maximal ideal

    Show that every non-unit of \(R\) is contained in a maximal ideal.

    Solution

    This follows because if \(x\in R\sm \unitsof{R}\), then \(Rx \normal R\) and \(Rx\neq R\) implies \(R/Rx \neq 0\). Then there exists some \(\bar \mfm \in \mspec R/Rx\), and by the correspondence theorem this lifts to some \(\mfm \in \mspec R\) containing \(Rx\).

    (Existence statements proved by Zorn's lemma)

    Every nonzero ring has a maximal ideal, every proper ideal is contained in a maximal ideal, and every vector space has a basis. Each proof applies Zorn’s lemma to a poset of partial objects ordered by inclusion – proper ideals containing a given ideal, or linearly independent subsets – in which the union of a chain is again an element of the poset and is an upper bound for the chain.

    Nakayama’s lemma

    Let \((R,\mfm)\) be a local ring and \(M\) a finitely generated \(R\)-module. If \(\mfm M=M\), then \(M=0\). Equivalently, elements \(m_1,\ldots,m_n\in M\) generate \(M\) if and only if their images span the \(R/\mfm\)-vector space \(M/\mfm M\).

    Nakayama's lemma

    If $M$ is finitely generated, $IM=M$, and $I\subseteq J(R)$, then $M=0$.Suggest a slogan

    Let \(R\) be a commutative ring, let \(I\subseteq R\) be an ideal, and let \(M\) be a finitely generated \(R\)-module with \(IM=M\). Then there is \(i\in I\) with \(im=m\) for every \(m\in M\). In particular, if \(I\) is contained in the Jacobson radical \(J(R)\), then \(M=0\).

    Let \((R,\mfm)\) be a local ring, let \(M\) and \(N\) be \(R\)-modules with \(N\) finitely generated, and let \(\varphi\colon M\to N\) be \(R\)-linear. Then \(\varphi\) is surjective if and only if the induced map \(M/\mfm M\to N/\mfm N\) of \(R/\mfm\)-vector spaces is surjective.

    Let \(m_1,\ldots,m_n\) generate \(M\). Since \(M=IM\), there are \(a_{jk}\in I\) with \(m_j=\sum_ka_{jk}m_k\), so the matrix \(\mathrm{I}_n-A\), with \(A=(a_{jk})\), kills the column vector \((m_1,\ldots,m_n)^t\). Multiplying by the adjugate of \(\mathrm{I}_n-A\) gives \(\det(\mathrm{I}_n-A)\,m_j=0\) for every \(j\). Expanding the determinant, \(\det(\mathrm{I}_n-A)=1-i\) with \(i\in I\), so \(m=im\) for every \(m\in M\). If \(I\subseteq J(R)\), then \(1-i\) is a unit by FF-QWCKR, so \(M=0\).

    For the local statement, only the converse needs proof. If \(M/\mfm M\to N/\mfm N\) is surjective, then \(N=\varphi(M)+\mfm N\), so the finitely generated module \(C=N/\varphi(M)\) satisfies \(\mfm C=C\), and \(C=0\) by the first part with \(I=\mfm=J(R)\).

    What is Nakayama's lemma?

    Nakayama's lemma for a local ring

    Minimal generators over a local ring are exactly bases modulo the maximal ideal.Suggest a slogan

    Let \((R,\mfm)\) be a local ring with residue field \(k=R/\mfm\), and let \(M\) be a finitely generated \(R\)-module. Elements \(m_1,\ldots,m_n\in M\) generate \(M\) if and only if their images span the \(k\)-vector space \(M/\mfm M\). Consequently, a generating set of \(M\) is minimal under inclusion if and only if its image is a \(k\)-basis of \(M/\mfm M\), and every minimal generating set of \(M\) has \(\dim_k M/\mfm M\) elements.

    If \((m_1,\ldots,m_n)\) and \((m_1',\ldots,m_n')\) are minimal generating sets of \(M\), then there is \(P\in\GL_n(R)\) with \(m_i'=\sum_jP_{ij}m_j\) for every \(i\).

    If the images of \(m_1,\ldots,m_n\) span \(M/\mfm M\), let \(N=Rm_1+\cdots+Rm_n\). Then \(M=N+\mfm M\), so \(\mfm(M/N)=M/N\), and \(M/N=0\) by Nakayama's lemma, since \(M/N\) is finitely generated and \(\mfm=J(R)\). The converse is immediate.

    If the image of a generating set is linearly dependent, some element has image in the span of the images of the others, and those others still generate \(M\) by the first part; if the image is a basis, no proper subset has spanning image.

    For the last claim, write \(m_i'=\sum_jP_{ij}m_j\) and \(m_j=\sum_lQ_{jl}m_l'\) with \(P,Q\in\Mat_n(R)\). Reducing modulo \(\mfm\), the matrices \(\overline P\) and \(\overline Q\) are the change-of-basis matrices between two bases of \(M/\mfm M\), so \(\overline P\,\overline Q=I\). Hence \(\det P\notin\mfm\), so \(\det P\) is a unit of \(R\) and \(P\in\GL_n(R)\).

    What is Nakayama's lemma for a local ring?

    Finitely generated flat modules over a Noetherian local ring are free

    Over a Noetherian local ring, finitely generated flat means free.Suggest a slogan

    Let \((R,\mfm)\) be a Noetherian local ring and let \(M\) be a finitely generated flat \(R\)-module. Then \(M\) is free.

    Let \(k=R/\mfm\) and \(n=\dim_kM/\mfm M\). By Nakayama's lemma for a local ring, lifts of a \(k\)-basis of \(M/\mfm M\) give a surjection \(R^n\to M\); let \(K\) be its kernel, which is finitely generated because \(R\) is Noetherian. Since \(M\) is flat, \(\Tor_1^R(k,M)=0\), so applying \(k\otimes_R-\) to \(0\to K\to R^n\to M\to0\) gives an exact sequence \[ 0\to K/\mfm K\to k^n\to M/\mfm M\to0. \] The map \(k^n\to M/\mfm M\) is a surjection between \(k\)-vector spaces of dimension \(n\), hence an isomorphism, so \(K/\mfm K=0\). By Nakayama's lemma, \(K=0\), and \(M\cong R^n\).

    Applied to stalks, this shows that a coherent sheaf \(\mathcal F\) on a Noetherian scheme \(X\) whose stalks \(\mathcal F_x\) are flat \(\OO_{X,x}\)-modules has free stalks, and is therefore locally free, that is, the sheaf of sections of a vector bundle.

    Give a categorical/homological corollary of Nakayama's lemma.

    Geometric interpretation of Nakayama's lemma

    Fiber bases lift to local generators, and to local frames for vector bundles.Suggest a slogan

    Let \(X\) be a Noetherian scheme, let \(\mathcal F\) be a coherent \(\OO_X\)-module, let \(p\in X\), and let \(\kappa(p)=\OO_{X,p}/\mfm_p\). The fiber \(\mathcal F(p)\coloneqq\mathcal F_p/\mfm_p\mathcal F_p\) of \(\mathcal F\) at \(p\) is a finite-dimensional \(\kappa(p)\)-vector space.

    1. Every lift to the stalk \(\mathcal F_p\) of a basis of \(\mathcal F(p)\) is a minimal generating set of \(\mathcal F_p\).

    2. Given such a lift, there are an open neighborhood \(U\) of \(p\) and sections \(s_1,\ldots,s_n\in\mathcal F(U)\) with these germs at \(p\) that generate \(\mathcal F_q\) for every \(q\in U\).

    3. If \(\mathcal F\) is locally free near \(p\), the sections \(s_1,\ldots,s_n\) of (b) can be chosen so that \(\OO_U^n\to\mathcal F|_U\), \((f_i)\mapsto\sum_if_is_i\), is an isomorphism; that is, a basis of the fiber of a vector bundle at \(p\) extends to a local frame on a neighborhood of \(p\).

    The stalk \(\mathcal F_p\) is a finitely generated module over the local ring \(\OO_{X,p}\), so (a) is Nakayama's lemma for a local ring.

    For (b), choose an open neighborhood \(V\) of \(p\) on which all the germs are represented by sections \(s_i\in\mathcal F(V)\), and let \(\mathcal C\) be the cokernel of \(\OO_V^n\to\mathcal F|_V\). Then \(\mathcal C\) is coherent with \(\mathcal C_p=0\), and the support of a coherent sheaf is closed, so \(\mathcal C\) vanishes on an open neighborhood \(U\subseteq V\) of \(p\).

    For (c), shrink \(U\) to an affine open \(\Spec A\) containing \(p\) on which \(\mathcal F\) is free; its rank is \(\dim_{\kappa(p)}\mathcal F(p)=n\). Since \(\OO_U^n\to\mathcal F|_U\) is surjective and \(U\) is affine, the map \(A^n\to\mathcal F(\Spec A)\cong A^n\) is a surjective endomorphism of a finitely generated \(A\)-module, hence an isomorphism.

    What is the geometric interpretation of Nakayama's lemma?

    Closed immersion criterion for morphisms of smooth projective varieties

    Over an algebraically closed field, smooth projective maps are closed immersions exactly when they are injective on closed points and tangent spaces.Suggest a slogan

    Let \(k\) be an algebraically closed field and \(f\colon X\to Y\) a morphism of smooth projective varieties over \(k\). Then \(f\) is a closed immersion if and only if \(f\) is injective on closed points and the differential \(df_p\colon T_pX\to T_{f(p)}Y\) is injective for every closed point \(p\in X\) [@Har10a].

    The application of Nakayama’s lemma is in the local step: a local homomorphism \(\mathcal O_{Y,f(p)}\to\mathcal O_{X,p}\) of Noetherian local rings that is an isomorphism on residue fields, surjective on cotangent spaces and makes \(\mathcal O_{X,p}\) a finitely generated module is surjective [@Har10a].

    Give a geometric application of Nakayama's lemma.

    Noetherian rings and Krull dimension

    A ring is Noetherian if every ascending chain of ideals stabilizes, equivalently if every ideal is finitely generated. Let \(R\) be a Noetherian ring and \(I\subseteq R\) an ideal.

    • By Krull’s principal ideal theorem, a prime ideal minimal over a principal ideal has height at most \(1\).

    • By the Artin–Rees lemma, for a finitely generated \(R\)-module \(M\) and a submodule \(N\subseteq M\) there is \(c\geq 0\) with \(I^nM\intersect N = I^{n-c}(I^cM\intersect N)\) for all \(n\geq c\).

    • By Krull’s intersection theorem, if \(R\) is local and \(I\) is proper, then \(\Intersect_{n\geq 1}I^nM=0\) for every finitely generated \(R\)-module \(M\).

    Krull's Hauptidealsatz

    In a Noetherian ring, a principal nonzerodivisor cuts codimension one.Suggest a slogan

    Let \(R\) be a commutative Noetherian ring and let \(a\in R\) be neither a zero divisor nor a unit. Then every prime ideal \(\mfp\) minimal among the prime ideals containing \(a\) has height \(1\).

    What is Krull's Hauptidealsatz?

    Krull's intersection theorem

    In a Noetherian ring, powers of a Jacobson-radical ideal have zero intersection.Suggest a slogan

    Let \(R\) be a commutative Noetherian ring and let \(I\subseteq R\) be an ideal contained in the Jacobson radical \(J(R)\). Then \[ \bigcap_{n\ge0}I^n=0. \] In particular, \(\bigcap_{n\ge0}\mfm^n=0\) for every Noetherian local ring \((R,\mfm)\).

    Let \(N=\bigcap_{n\ge0}I^n\). By the Artin--Rees lemma applied to \(N\subseteq R\), there is \(c\ge0\) with \(I^n\cap N=I^{n-c}(I^c\cap N)\) for all \(n\ge c\). Since \(N\subseteq I^n\) for every \(n\), taking \(n=c+1\) gives \(N=I(I^c\cap N)=IN\). The ideal \(N\) is finitely generated because \(R\) is Noetherian, so Nakayama's lemma with \(I\subseteq J(R)\) gives \(N=0\).

    What is Krull's intersection theorem?

    Geometric interpretation of Krull's intersection theorem

    On a locally Noetherian scheme, a nonzero germ cannot vanish to infinite order.Suggest a slogan

    Let \(X\) be a locally Noetherian scheme, for example a variety, let \(p\in X\), and let \((\OO_{X,p},\mfm_p)\) be the local ring at \(p\). For \(f\in\OO_{X,p}\), define the order of vanishing of \(f\) at \(p\) as \[ \ord_p(f)\coloneqq\sup\theset{n\ge0\suchthat f\in\mfm_p^n}\in\ZZ_{\ge0}\cup\theset{\infty}. \] Then \(\ord_p(f)=\infty\) if and only if \(f=0\): a nonzero function vanishes to finite order at \(p\).

    The local ring \(\OO_{X,p}\) is Noetherian, and \(\ord_p(f)=\infty\) means \(f\in\bigcap_{n\ge0}\mfm_p^n\), which is \(0\) by Krull's intersection theorem.

    Nakayama's lemma for a local ring gives the companion statement for generators. Let \(k\) be a field and let \(\mfm_0\subseteq\OO_{\AA^n_k,0}\) be the maximal ideal at the origin. Functions \(f_1,\ldots,f_n\in\mfm_0\) generate \(\mfm_0\), that is, form a local coordinate system at \(0\), if and only if their classes form a basis of the cotangent space \(\mfm_0/\mfm_0^2\). The class of \(f\) in \(\mfm_0/\mfm_0^2\) is its linear part, the differential \(df_0=\sum_i\frac{\partial f}{\partial x_i}(0)\,dx_i\), so the condition is that the Jacobian matrix \(\qty{\frac{\partial f_i}{\partial x_j}(0)}\) is invertible, as in the inverse function theorem.

    What is the geometric interpretation of Krull's intersection theorem?

    The Artin--Rees lemma

    Deep enough in an $I$-adic filtration, intersection with a submodule is a fixed shifted $I$-adic filtration.Suggest a slogan

    Let \(R\) be a commutative Noetherian ring, let \(I\subseteq R\) be an ideal, let \(M\) be a finitely generated \(R\)-module, and let \(N\subseteq M\) be a submodule. Then there is an integer \(c\ge0\) such that for every \(n\ge c\), \[ I^nM\cap N=I^{n-c}\qty{I^cM\cap N}. \]

    What is the Artin-Rees lemma?

    Integral extensions

    For an integral extension of rings \(A\subseteq B\), every prime of \(A\) is the contraction of a prime of \(B\), and by the going-up theorem a chain of primes \(\mfp_1\subseteq\cdots\subseteq\mfp_n\) of \(A\) together with a prime \(\mathfrak q_1\) of \(B\) satisfying \(\mathfrak q_1\intersect A=\mfp_1\) extends to a chain \(\mathfrak q_1\subseteq\cdots\subseteq\mathfrak q_n\) of primes of \(B\) with \(\mathfrak q_i\intersect A=\mfp_i\).

    Lying over and going up for integral extensions

    Integral extensions make spectra surjective and lift ascending chains of primes.Suggest a slogan

    Let \(R\subseteq S\) be an integral extension of commutative rings.

    1. Lying over: for every prime ideal \(\mfp\subseteq R\) there is a prime ideal \(\mathfrak q\subseteq S\) with \(\mathfrak q\cap R=\mfp\). Equivalently, the map \(\Spec S\to\Spec R\), \(\mathfrak q\mapsto\mathfrak q\cap R\), is surjective.

    2. Going up: if \(\mfp_1\subseteq\mfp_2\) are prime ideals of \(R\) and \(\mathfrak q_1\) is a prime ideal of \(S\) with \(\mathfrak q_1\cap R=\mfp_1\), then there is a prime ideal \(\mathfrak q_2\supseteq\mathfrak q_1\) of \(S\) with \(\mathfrak q_2\cap R=\mfp_2\).

    What is the going up theorem?

    Localization

    Field of fractions

    Let \(R\) be an integral domain. On \(R \times (R\setminus\theset{0})\) define \((a,s)\sim (b, t)\) if \(at = bs\), and write \(a/s\) for the class of \((a,s)\). The field of fractions of \(R\) is the set of classes \[ \ff(R) \coloneqq \bigl(R \times (R\setminus\theset{0})\bigr)/{\sim}, \] with operations \(\frac as+\frac bt=\frac{at+bs}{st}\) and \(\frac as\cdot\frac bt=\frac{ab}{st}\).

    Transitivity of \(\sim\) uses that \(R\) has no nonzero zero divisors: if \(at=bs\) and \(bu=ct\), then \(aut=bsu=cst\), so \(t(au-cs)=0\), and \(t\neq0\) gives \(au=cs\). The same hypothesis gives \(st\neq 0\), so the operations are defined on \(R\times(R\setminus\theset{0})\).

    For a submonoid \(S\leq (R,\cdot)\), write \(S^{-1}R\) for the localization of \(R\) at \(S\); its construction and universal property are in the Stacks Project, Tag 00CM.

    The canonical map \[ \begin{aligned} R &\to R\localize{S} \\ x &\mapsto {x\over 1} \end{aligned} \] need not be injective: its kernel is \(\ts{x\in R \st sx=0 \text{ for some } s\in S}\).

    For an integral domain \(R\), \[ \ff(R) \cong R\localize{ (\nonzeroof{R}) }. \]

    Hilbert’s basis theorem and primary ideals

    Hilbert basis theorem

    Adjoining finitely many polynomial variables preserves Noetherianity.Suggest a slogan

    If \(R\) is a commutative Noetherian ring, then the polynomial ring \(R[x]\) is Noetherian. By induction, \(R[x_1, \ldots, x_n]\) is Noetherian for every \(n\); in particular every ideal of \(k[x_1, \ldots, x_n]\) over a field \(k\) is finitely generated, and every ideal of \(\ZZ[x_1,\ldots,x_n]\) is finitely generated.

    [@DF04].

    What is the Hilbert basis theorem?

    Primary ideal

    Let \(R\) be a commutative ring. A proper ideal \(I\subsetneq R\) is primary if for all \(p,q\in R\) with \(pq\in I\), either \(p\in I\) or there exists \(n\geq 1\) such that \(q^n\in I\).

    In a commutative ring, every maximal ideal is prime, and every prime ideal is both radical and primary. [@AM18]

    In \(k[x,y]\) the ideal \((xy)\) is radical but not primary, since \(x \cdot y \in (xy)\) while \(x \notin (xy)\) and no power of \(y\) lies in \((xy)\). In \(k[x]\) the ideal \((x^2)\) is primary but not radical, since \(x \notin (x^2)\) and \(x^2 \in (x^2)\).

    Show that maximal implies prime implies both radical and primary, and that neither radical nor primary implies the other.

    If \(R\) is Noetherian, then so is \(R[x_1,\ldots,x_n]\), by Hilbert’s basis theorem and induction on \(n\). An ideal \(\mathfrak q\subsetneq R\) is primary if every zero divisor of \(R/\mathfrak q\) is nilpotent; every prime ideal is primary, and by the Lasker–Noether theorem every ideal of a Noetherian ring is a finite intersection of primary ideals.

    What links to this