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    Commutative algebra

    For a multiplicative submonoid \(S\) of a commutative ring \(R\), the localization \(S^{-1}R\) is the initial ring under \(R\) in which every element of \(S\) is a unit. For a prime ideal \(\mathfrak p\), \(R_\mathfrak p=(R\setminus\mathfrak p)^{-1}R\) is a local ring with maximal ideal \(\mathfrak pR_\mathfrak p\). Nakayama’s lemma: for a finitely generated module \(M\) over a local ring \((R,\mathfrak m)\), \(\mathfrak mM=M\) implies \(M=0\). The nilradical of \(R\), the set of nilpotent elements, is the intersection of all prime ideals; the Jacobson radical is the intersection of all maximal ideals; and for a finitely generated algebra over a field the two coincide. A ring is Noetherian if every ideal is finitely generated, and by Hilbert’s basis theorem \(R[x]\) is Noetherian when \(R\) is Noetherian.

    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 \(R\times(R\setminus\theset{0})\).

    Local ring

    A commutative ring \(R\) is local if it has a unique maximal ideal \(\mfm\), that is, \(\mspec R = \theset{ \mfm }\). The residue field of the local ring \((R,\mfm)\) is \(\kappa \coloneqq R/\mfm\).

    Regular local rings, regular rings, and DVRs

    A Noetherian local ring \((A,\mm)\) is a regular local ring if \[ \dim A = \dim_{A/\mm} \mm/\mm^2 . \] A Noetherian ring \(A\) is a regular ring if \(A_\mfp\) is a regular local ring for every prime \(\mfp\).

    (valuation)

    Let \(K\) be a field and \(G\) a totally ordered abelian group. A valuation of \(K\) with values in \(G\) is a group homomorphism \(v \colon K^\times \to G\) such that \(v(x+y) \geq \min(v(x), v(y))\) for all \(x, y \in K^\times\) with \(x + y \neq 0\). Its valuation ring is \(R = \ts{x \in K^\times \st v(x) \geq 0} \cup \ts{0}\). A domain \(R\) is a valuation ring if it is the valuation ring of some valuation of \(\operatorname{Frac} R\), and the valuation is discrete if \(G = \ZZ\). [@Har10a, §I.6]

    By Krull’s height theorem, \(\dim A \leq \dim_{A/\mm}\mm/\mm^2\) for every Noetherian local ring, and \(\dim_{A/\mm}\mm/\mm^2\) is the minimal number of generators of \(\mm\). If \(A\) is regular, then \(\mm\) is generated by a regular sequence, and \(A\) is Cohen–Macaulay.

    A regular local ring is a domain, is a UFD by the Auslander–Buchsbaum theorem, and is therefore integrally closed. By Serre’s theorem, a Noetherian local ring is regular if and only if its residue field has finite projective dimension, if and only if it has finite global dimension; consequently every localization of a regular local ring at a prime is regular.

    For a Noetherian local domain of dimension one, the following are equivalent: it is regular; it is a DVR; it is integrally closed; \(\mm\) is principal. Hence on a Noetherian integral scheme regular in codimension one, the local ring at the generic point of each prime divisor \(Y\) is a DVR, and its valuation \(v_Y\) gives the coefficients of the principal Weil divisor \(\operatorname{div}(f)\) (D-5PQ5W).

    When is a Noetherian local ring regular?
    Characterise discrete valuation rings among Noetherian local rings.
    Is a regular local ring a domain? Is it integrally closed?
    Show that $R$ is a discrete valuation ring if and only if $R$ is a local PID that is not a field, and define a uniformizer.
    Show that the ring of germs at $0$ of holomorphic functions on a disc is a discrete valuation ring with uniformizer $z$.
    For the local ring $R$ of a curve at a point, over $k = \bar k$, identify the closed and generic points of $\Spec R$ and their residue fields, and decide which maps between $\Spec R$ and $\Spec \operatorname{Frac} R$ are morphisms of schemes.

    Regular local rings and regular rings

    Regular local rings have no excess generators: $\dim_k \mfm/\mfm^2=\dim R$.Suggest a slogan

    A Noetherian local ring \((R,\mfm)\) with residue field \(k=R/\mfm\) is regular if \(\mfm\) can be generated by \(\dim R\) elements, where \(\dim R\) is the Krull dimension of \(R\). By Nakayama's lemma for a local ring, the minimal number of generators of \(\mfm\) is \(\dim_k\mfm/\mfm^2\), so \(R\) is regular if and only if \(\dim_k\mfm/\mfm^2=\dim R\).

    A commutative Noetherian ring \(R\) is regular if the localization \(R_{\mfp}\) is a regular local ring for every prime ideal \(\mfp\subseteq R\).

    By Krull’s height theorem, the extension of Krull's Hauptidealsatz to ideals generated by \(r\) elements, \(\dim R\le\dim_k\mfm/\mfm^2\) for every Noetherian local ring \((R,\mfm)\); regularity is the case of equality.

    What is a regular ring?

    Valuation ring

    An integral domain \(R\) with field of fractions \(K\) is a valuation ring if for every nonzero \(x\in K\), \(x\in R\) or \(x^{-1} \in R\).

    Discrete valuation ring

    A discrete valuation ring, or DVR, is a principal ideal domain that is a local ring and is not a field.

    Noetherian ring

    A ring \(R\) is Noetherian if it satisfies the ascending chain condition on ideals: for every chain of ideals \(I_1 \subseteq I_2 \subseteq \cdots\) of \(R\) there exists \(N\geq 1\) such that \(I_N = I_{N+1} = \cdots\).

    Noetherian module

    Let \(R\) be a ring. An \(R\)-module \(M\) is Noetherian if there is no infinite strictly increasing sequence \(M_1 \subsetneq M_2 \subsetneq \cdots\) of submodules of \(M\).

    What does it mean for a module to be Noetherian?

    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?

    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?

    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?

    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?

    Nilpotent element

    Let \(R\) be a ring. An element \(r\in R\) is nilpotent if \(r^n = 0\) for some integer \(n \geq 1\).

    Nilradical

    Let \(R\) be a commutative ring. The nilradical of \(R\) is the set of nilpotent elements of \(R\): \[ \nilrad{R} \coloneqq \theset{x\in R \suchthat x \text{ is nilpotent}}. \]

    Radical of an ideal

    Let \(R\) be a commutative ring and \(I\subseteq R\) an ideal. The radical of \(I\) is \[ \rad{I} \coloneqq \theset{r\in R\suchthat r^n\in I\text{ for some } n\geq 0}. \] The ideal \(I\) is radical if \(\rad{I} = I\).

    Reduced ring

    A ring \(R\) is reduced if \(R\) contains no nonzero nilpotent elements.

    Reduced ring

    Reduced means no nonzero nilpotents.Suggest a slogan

    A commutative ring \(R\) is reduced if and only if it has no nonzero nilpotent elements, that is, its nilradical \(\nilrad{R}\) is \(0\).

    What is a reduced ring?

    Jacobson radical

    Let \(R\) be a commutative ring. The Jacobson radical of \(R\) is the intersection of all maximal ideals of \(R\): \[ \jacobsonrad{R} \coloneqq \Intersect_{\mfm \in \mspec R} \mfm, \] where \(\mspec R\) is the max spectrum of \(R\).

    The Jacobson radical $J(R)$

    $J(R)$ annihilates every simple module, and for finite-type algebras over a field it is the nilradical.Suggest a slogan

    Let \(R\) be a commutative ring. The Jacobson radical of \(R\) is the intersection of the annihilators of the simple \(R\)-modules: \[ J(R)=\bigcap_{\mfm\in\mspec R}\mfm=\bigcap_{M\text{ simple}}\Ann_R(M). \] If \(R\) is a finitely generated algebra over a field \(k\), then \(J(R)\) equals the nilradical \(\nilrad{R}\).

    A simple module \(M\) is generated by any nonzero \(m\in M\), so \(M\cong R/\Ann_R(m)\), and simplicity makes \(\Ann_R(m)\) a maximal ideal \(\mfm\); then \(\Ann_R(M)=\mfm\). Conversely \(R/\mfm\) is simple with annihilator \(\mfm\) for every maximal ideal \(\mfm\). So the two intersections run over the same ideals.

    For a finitely generated \(k\)-algebra, every prime ideal is an intersection of maximal ideals by the Nullstellensatz, so the intersection of all maximal ideals equals the intersection of all prime ideals, which is \(\nilrad{R}\).

    Write \(R=k[x_1,\ldots,x_n]/I\). Then \(J(R)=\nilrad{R}=\sqrt I/I\), so \(J(R)=0\) exactly when \(I\) is a radical ideal; for \(k\) algebraically closed, these are the ideals \(I(Z)\) of affine algebraic sets \(Z\subseteq\AA^n_k\), whose coordinate rings \(k[x_1,\ldots,x_n]/I(Z)\) are the rings of regular functions on \(Z\).

    What is ${J ({R}) }$? How is it interpreted geometrically?

    Quasiregular elements and the Jacobson radical

    The Jacobson radical is the largest ideal of quasiregular elements.Suggest a slogan

    Let \(R\) be a commutative ring. An element \(a\in R\) is quasiregular if \(1-a\) is a unit of \(R\). The Jacobson radical \(J(R)\) is the largest ideal of \(R\) all of whose elements are quasiregular: every element of \(J(R)\) is quasiregular, and every ideal consisting of quasiregular elements is contained in \(J(R)\).

    Let \(x\in J(R)\). If \(1-x\) were not a unit, it would lie in a maximal ideal \(\mfm\) by FF-QELG7, and so would \(x\), giving \(1\in\mfm\).

    Let \(I\) be an ideal consisting of quasiregular elements, let \(x\in I\), and suppose \(x\notin\mfm\) for a maximal ideal \(\mfm\). Then \(\mfm+(x)=R\), so \(1=y+rx\) with \(y\in\mfm\) and \(r\in R\). Since \(rx\in I\), the element \(y=1-rx\) is a unit, contradicting \(y\in\mfm\). So \(I\subseteq\mfm\) for every maximal ideal \(\mfm\), that is, \(I\subseteq J(R)\).

    What is a quasiregular element in a ring?

    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.

    Irreducible ideal

    Let \(R\) be a commutative ring. An ideal \(I\normal R\) is irreducible if for all ideals \(J_1, J_2 \normal R\) with \(I = J_1 \cap J_2\), either \(J_1 = I\) or \(J_2 = I\).

    $\gens{p}$-primary module

    Let \(R\) be a principal ideal domain, \(p\in R\) a prime, and \(M\) an \(R\)-module. \(M\) is \(\gens{p}\)-primary if every \(m\in M\) is annihilated by some power of \(p\): for each \(m\in M\) there is \(n\geq 1\) with \(p^n m = 0\).

    For any \(R\)-module \(M\), the \(p\)-primary component \(M[p^\infty] \coloneqq \theset{ m \in M \st p^n m = 0 \text{ for some } n\geq 1}\) is a submodule, and \(M\) is \(\gens{p}\)-primary if and only if \(M[p^\infty]=M\).

    For $R$ a PID and $p \in R$ prime, when is an $R\dash$module $\gens{p}\dash$primary?

    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.

    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?

    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?

    Gorenstein ring

    A Noetherian local ring \(R\) is Gorenstein if \(R\), as a module over itself, has finite injective dimension. A commutative Noetherian ring \(R\) is Gorenstein if its localization \(R_{\mathfrak p}\) is a Gorenstein local ring for every prime ideal \(\mathfrak p\subseteq R\) [@Eis95].

    A commutative Noetherian ring of finite injective dimension over itself is Gorenstein; conversely, a Gorenstein ring of finite Krull dimension has finite injective dimension over itself [@Eis95].