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\title{Lecture 5-29: Review, part 2}
\date{May 29, 2024}
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We continue with review, now summarizing the material on group cohomology and starting on algebraic geometry.
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We start with the notion of projective module.  Given a ring $R$ (not necessarily commutative) and a left $R$-module $M$, we say that $M$ is \alert{projective} if given any surjection $g:N\rightarrow N'$ of left $R$-modules and an $R$-module map $f:M\rightarrow N'$ there is an $R$-module map $h:M\rightarrow N$ with $gh=f$ (so that $h$ lifts $f$ to $N$).  It is easy to show that $M$ is projective if and only if $M$ is a direct summand of a free $R$-module.  Thus, in particular, if $R$ is a PID and $M$ is finitely generated (or even in general), $M$ is projective if and only if $M$ is free.
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 In particular,  every free $R$-module is  projective; since any $R$-module $A$ is a homomorphic image of a free module (for example, the one with any set of generators of $A$ as a free basis) one sees that any such module $A$ admits a \alert{projective resolution} $\cdots P_n\rightarrow\cdots\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow A\rightarrow 0$, so that the $P_i$ are projective and the sequence is exact.  Taking homomorphisms into an $R$-module $D$, we get a cochain complex $0\rightarrow\hom_R(A,D)\rightarrow\hom_R(P_0,D)\rightarrow\cdots$, where $d_n:\hom_R(P_{n-1},D)\rightarrow\hom_R(P_n,D)$.  The the $n$th cohomology group of this complex, or the quotient $\ker d_{n+1}/$ im $d_n$, is defined to be the $n$th Ext group Ext$^n_R(A,D)$ of $A$ (with coefficients in $D$).  It is independent of the choice of projective resolution $\{P_i\}$. In particular Ext$^0_R(A,D)=\hom_R(A,D)$.  The Ext groups are $\mathbb Z$-modules, but not $R$-modules, unless $R$ is commutative.
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The cohomology groups $H^n(G,A)$ of a (finite) group $G$ with coefficients in a $G$-module $A$ are a special case of Ext groups.  More precisely, let $R=\mathbb ZG$ be the \alert{integral group ring} consisting of all combinations $\sum\limits_{g\in G}z_gg$ with $z_g\in\mathbb Z$ and addition and multiplication defined in the obvious way.  A $G$-module $A$ is then just a (left) $R$--module; in particular, one has the \alert{trivial $G$-module} $\mathbb Z$, on which $G$ acts trivially.  Then the cohomology group $H^n(G,A)$ is defined to be Ext$_R^n(\mathbb Z,A)$.  In particular $H^0(G,A)$ identifies with $A^G$, the subgroup of $G$-fixed elements in $A$.
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 If $G=\mathbb Z_m$ is cyclic of order $m$, say with generator $\sigma$, then one has a simple explicit resolution of $\mathbb Z$ in which every term apart from $\mathbb Z$ is $R=\mathbb ZG$; the maps toggle between multiplication by $N=1+\sigma+\cdots+\sigma^{m-1}$ and multiplication by $\sigma-1$, except for the last one, which sends $\sum z_gg$ to $\sum z_g$.  It follows easily that $H^0(G,A)\cong A^G,H^n(G,A)\cong A^G/NA$ if $n$ is even and $n\ge2$, while $H^n(G,A)\cong{}_NA/(\sigma-1)A$ if $n$ is odd, where ${}_NA=\{a\in A:Na=0\}$.  For general groups $G$, there is a standard resolution called the bar resolution to compute $H^n(G,A)$; but don't worry about the details of this resolution.
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An easy consequence of the definition of $H^n(G,A)$ is that \alert{if $A$ is torsion, so that there is a nonzero integer $m$ with $mA=0$, then we also have $mH^n(G,A)=0$ for all $n$}.  A deeper fact, relying on maps called the \alert{restriction} and \alert{corestriction} maps, is that \alert{$|G|H^n(G,A)=0$ for all $n>0$ if $G$ has order $|G|$}.  Putting these two constraints together, we deduce that \alert{$H^n(G,A)=0$ for all $n>0$ if $A$ and $G$ are finite with relatively prime orders}.  As a consequence of this vanishing result together with the interpretation of $H^2(G,A)$ by group extensions, we obtain \alert{Schur's Theorem} that \alert{any finite group $E$ admitting a normal subgroup $A$ (not necessarily abelian) whose index is relatively prime to its order necessarily has a subgroup $G$ isomorphic to $E/A$, so that $E$ is the semidirect product of $A$ and $G$}.  Moreover, \alert{any two complements of $A$ in $E$ are conjugate under $A$ if $A$ is abelian}.  
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Turning now to algebraic geometry, we start with an algebraically closed field $k$ and recall that in the \alert{Zariski topology} the closed, or algebraic, subsets of $k^n$ (now denoted $\mathbb A^n$ and called \alert{affine space}) are precisely the zero loci $\mathcal Z(I)$ of ideals $I$ in the polynomial ring $P_n=k[x_1,\ldots,x_n]$.  The Nullstellensatz then asserts the existence of inclusion-reversing inverse bijections between radical ideals in $P_n$ and closed subsets of $\mathbf A^n$, mapping an ideal $I$ to its zero locus $\mathcal Z(I)$ and a closed set $Z$ to the ideal  $I(Z)$ of polynomials vanishing on $Z$.  
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A closed subset $V$ is called \alert{irreducible} if it is not the union of two (nonempty) proper closed subsets.  Any closed set is then uniquely the union of finitely many irreducible closed subsets, none contained in another; these can overlap, unlike the connected components of a topological space.  We call $V$ a \alert{variety} if it is irreducible.   Subvarieties of $\mathbf A^n$ then correspond to prime ideals in $P_n$ via the Nullstellensatz bijection.  
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The Nullstellensatz was proved via the \alert{Noether normalization lemma}, which asserts for any field $k$ that \alert{any finitely generated $k$-algebra is a finitely generated integral extension of some polynomial ring $P_d$ over $k$}.  Using this we showed that the only finitely generated $k$-algebras that are field are the finite extensions of $k$, from which one can deduce a version of the Nullstellensatz that holds over any field $k$.  Also Noether normalization gives a very convenient alternate characterization of the dimension of a variety $V$ (that is, an irreducible closed subset of $\mathbf A^n$).  The dimension of $V$ was initially defined as the transcendence degree of the quotient field $k(V)$ of its coordinate ring $k[V]=P_n/I(v), I$ the ideal of $V$; but it can also be defined as the integer $d$ such that $k[V]$ is integral over $P_d$ (this integer being unique).
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Morphisms $\pi$ (that is, roughly speaking, polynomial maps) from oneclosed set $V$ to another one $W$, not necessarily of the same affine space $\mathbf A^n$, correspond to $k$-algebra homomorphisms $\tilde\pi$ from the coordinate ring $k[W]$ to $k[V]$.  If $\pi$ is surjective then $\tilde\pi$ is injective; if $\pi$ is surjective then $\tilde\pi$ is surjective.  In particular any proper subvariety $W$ of a variety $V$ is such that $k[W]$ is a quotient of $k[V]$.  The dimension $\dim W$ must be strictly less than $\dim V$ in this situation.  
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More algebraically and generally, we can define the (Krull) dimension of a commutative ring $R$ to be the largest number $n$ such that there is a chain $P_0\subset\cdots\subset P_n$ of distinct prime ideals in $R$.  The dimension of an algebraic set $V$ is then equal to the Krull dimension of its coordinate ring $k[V]$.
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Given an algebraic set $V\subset\mathbf A^n$ and a point $v\in V$, the \alert{tangent space $T_vV$ of $V$ at $v$} is the subspace of $k^n$ defined by the linear polynomials $\sum\limits_{i=1}^n D_if(v)x_i$, where $D_i$ denotes partial differentiation with respect to the $i$th variable $x_i$ and the functions $f$ run through the elements of the ideal $I(V)$ of $V$, or equivalently (thanks to the product rule) just through a set of generators of $I$.  On a nonempty open subset of $V$, this space has dimension equal to that of $V$; we call any point $v$ for which this holds \alert{nonsingular}.  The set of \alert{singular} (i.e., not nonsingular) points forms a proper closed subset of $V$, of smaller dimension; the tangent space at any of those points has dimension larger than that of $V$.
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