ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES

ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES (1) Let k be an algebraically-closed field throughout this week. (2) The ideals of k [X 1,...,X n ] form a lattice (a) The radical ideals form a sublattice (b) The prime ideals for a sublattice of the radical ideals (c) Every radical ideal is the intersection of a finite number of prime ideals (3) Let E = k n, Euclidean space (a) The polynomials in k [X 1,...,X n ] are functions E k (b) Algebraic subsubsets of E are zeros of one or more functions (i) If S k [X 1,...,X n ],thenv (S) ={x E,s(x) =0all s S} (ii) If S T k [X 1,...,X n ] then V (T ) V (S) (iii) If J =(S) then V (J) =V (S) (iv) For every subset S there exists a finite set T S of polynomials such that V (S) =V (T ). (A) T is a finite set of generators for the ideal generated by S. (v) finite unions and and infinite intersections of algebraic subsets are algebraic subsets (vi) \ V (J k )=V ( P k J k) k (vii) V (J 1 ) V (J 2 )=V (J 1 J 2 ) (viii) V (0) = E, V (k [X 1,...,X n ]) = φ. (c) If W is an algebraic subset of E then I (W )={p k [X 1,...,X n ]:p(x) =0, all x W } (i) If W 1 W 2 are algebraic subsets of E then I (W 2 ) I (W 1 ) (ii) I (W 1 W 2 )=I(W 1 ) I (W 2 ) (iii) p I (W 1 )+I (W 2 )=I(W 1 W 2 ) (iv) I (W ) is a radical ideal (nullstellensatz) (v) V (I (W )) = W (vi) I (V (S)) = p (S), I (V (J)) = J.(nullstellensatz) (vii) W is a single point of E = k n iff I (W ) k [X 1,...,X n ] is a maximal ideal (nullstellensatz) (4) The radical ideals are in 1 1 correspondence with the algebraic subsets of E (5) An algebraic subset is irreducible if it is not the union of two strictly smaller algebraic subsets. Irreducible algebraic subsets are called subvarieties. (a) Otherwise the algebraic subset is reducible (b) Every algebraic subset is the union of a finite number of irreducible algebraic subsets (c) Every radical ideal is the intersection of a finite number of prime ideals (6) A topology on E is a collection of subsets of E, called closed subsets, satisfying certain conditions (a) The collection of algebraic subsets of E are the closed subsets for a topology on E called the Zariski topology. 1

2 ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES (b) In the Zariski topology, every infinite intersection of closed sets is a finite intersection (c) In the Zariski topology, an algebraic subset is irreducible in the topological sense (not the union of two proper closed subsets) iff it is an irreducible as an algebraic subset (not the union of two proper algebraic subsets). (i) Proof: algebraic subset = closed set (ii) Corollary: an algebraic subset is irreducible iff all non-empty open subsets are dense iff no two non-empty open subsets have empty intersection (iii) Example: in k 2 the algebraic subset W = V (X 1 X 2 )=V (X 1 ) V (X 2 ) has a non-empty, non-dense open subset U = W V (X 1 ).TheclosureofU is U = V (X 2 ). (d) The Zariski topology on k hasclosedsetsconsistingoffinite sets and all of k (e) Polynomials are continuous functions E k in the Zariski topology (but not the only ones). (7) In everything we have done so far, all algebraic subsets are algebraic subsets of E = k n and all ideals are ideals of k [X 1,...,X n ]. (a) Beware: generalization coming. (8) Remark: since our ground field is infinite, every polynomial represents a different function E k. (9) Useful abstraction: let E be a set and R a ring. The set of R-valued functions on E is the set of functions f : E R. (a) The R-valued functions on E is a ring if we use pointwise addition and multiplication. (10) A polynomial function on an algebraic subset W E is the restriction of a polynomial p : E k to W. (a) If p and q are polynomials, then they represent the same polynomial function on W iff p q I (W ). (b) So different polynomials can represent the same polynomial function on W. (c) Let Γ (W ) be the set of polynomial functions on W. (i) Γ (E) =k [X 1,...,X n ]. (ii) Γ (W ) is a subring of the ring of k-valued functions on W (iii) There is a natural map, restriction, mapping k [X 1,...,X n ] to the k-valued functions on W. The image is Γ (W ) and the kernal is I (W ). π 0 I (W ) k [X 1,...,X n ] Γ (W ) 0 (iv) Thus Γ (W ) = k [X 1,...,X n ] is a ring. I (W ) (v) Γ (W ) is a finitely generated k-algebra. (vi) Γ (W ) has no nilpotent elements (vii) Γ (W ) is an integral domain iff I (W ) is prime iff W is irreducible (viii) Γ (W ) is a field iff I (W ) is maximal iff W is a single point

ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES 3 (11) Every finitely-generated k-algebra without nilpotents is the ring of polynomial functions for some algebraic subset (a) Given k [x 1,...,x n ],define π : k [X 1,...,X n ] k [x 1,...,x n ].Then ker (π) is a radical ideal, and if W is its algebraic subset then Γ (W ) = k [x 1,...,x n ]. (12) Fix an algebraic subset W,of E = k n. (a) There are 1 1 correspondences between the following sets: (i) radical ideals of Γ (W ) (ii) radical ideals of k [X 1,...,X n ] containing I (V ) (iii) algebraic subsets of E contained in W (b) The restriction of the Zariski topology to W is called the Zariski topology on W. The closed sets are the subvarieties of W. (c) Polynomial functions in Γ (W ) are continuous functions W k in the Zariski topology (d) If X W then I W (X) ={p Γ (W ):p(x) =0, all x X} (i) I W (X) =π (I (X)). (ii) If X Y W then I W (Y ) I W (X). (iii) I W (X) is a radical ideal (e) If S Γ (W ) is an ideal, define V W (S) ={x W : p (x) =0all p S} (i) V W (S) =V π 1 (S) (ii) if S T Γ (W ) then V W (T ) V W (S) (iii) If X is an algebraic subset then V W (I W (X)) = X (iv) If S is a radical ideal I W (V W (S)) = S (f) If X W are algebraic subsets of E, then there is a natural map Γ (W ) Γ (X). Infact: 0 I W (X) Γ (W ) Γ (X) 0 so Γ (X) = Γ (W ) I W (X). (g) X is a single point of W iff I W (X) is a maximal ideal of Γ (W ) (i) Serre s bright idea: even if k is not algebraically closed, you can use the set of maximal ideals of a, finitely generated k-algebra without nilpotents A as the ring of polynomial functions on a topological space whose point set is the collection of maximal ideals of A. (ii) Grothendieck extended this idea to include all of the prime ideals of A in the point set. Only the maximal ideals are closed points; the other prime ideals are non-closed points in the set. (h) X is irreducible in W (not the union of two proper closed subsets = not the union of two proper algebraic subsets) iff I W (X) is a prime ideal of Γ (W ). (i) Example: in k [x, y] = k [X, Y ],considertheideal(x). Theringisthe (XY ) ring of polynomial functions on the two axes in k 2,and(x) is the ideal of one of the axes. (13) Let s do something very general, the apply the idea to algebraic subsets. (a) If X and Y are sets, define Γ (X, Y ) as the set of all functions from X to Y. (i) If Y = R is a ring then Γ (X i,r) is a ring and an R-algebra.

4 ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES (ii) In many contexts X and Y will have some special properties or structure, and we will only consider maps respecting that structure. (iii) For example, if X and Y are topological spaces, we may restrict Γ (X, Y ) to continuous maps (iv) If X is a manifold and Y = R we may restrict Γ (X, Y ) to differentiable functions. (v) In the following, X will be an algebraic set and Y = k and Γ (X, k) =Γ (X) will be the set of polynomial functions on X. (b) Suppose X 1,X 2 and Y are sets and f : X 1 X 2 is a function. Then f induces a function f : Γ (X 2,Y) Γ (X 1,Y) as follows: f (α) =α f f g (i) If X 1 X 2 X 3 then (g f) = g f (draw diagram) (ii) If Y = R isaringthenγ (X i,r) are R-algebras and f is an R-algebra homomorphism. (iii) If Γ (X i,y) is a set of special functions, we have to be sure that f maps special functions to special functions. (A) If X i,y are topological spaces and f is continuous, then f maps continuous functions to continuous functions because the composition of continuous functions is continuous. (iv) If Γ (X, k) means the polynomial functions on X only, then what kind of function f : X 1 X 2 has the property that f takes polynomial functions on X 2 into polynomial functions on X 1? (14) A polynomial map P : k n k m is an orders set P =(p 1,...,p m ) of polynomials p i k [X 1,...,X n ].Thatis,ifx k n then P (x) =y =(y 1,...,y m ) where y i = p i (x) =p i (x 1,...,x n ).Inotherwords,P k [X 1,...,X n ] m. (a) Fix a point y k m. Then P 1 (y) ={x k n : P (x) =y} is an algebraic subset of k n,possiblyempty. (b) If W is an algebraic subset of k m then P 1 (W )={x k n : P (x) W }.is an algebraic subset of k n,possiblyempty. (c) Example: P : k k 3, k (x) = x, x 2,x 3, the image is a twisted cubic. (d) A polynomial map P induces a k-algebra homomorphism P : k [Y 1,...,Y m ] k [X 1,...,X n ].Infact,P (Y i )=p i, and that s all you have to know. (i) Proof: let F (k n ) be the k-valued functions on k n.thenk[y 1,...,Y m ] is a subring of F (k m ) and k [X 1,...,X n ] is a subring of F (k n ). We know that P : F (k m ) F (k n ) is a k-algebra homomorphism (13.b.i), so it suffices to show that P maps polynomials topolynomials. Forthatitsuffices to show that P (Y i ) is a polynomial. But P (Y i )=Y i P = p i. (e) Corollary: if k n P k m Q k t then (Q P ) = Q P (f) Every homomorphism ϕ : k [Y 1,...,Y m ] k [X 1,...,X n ] comes from a polynomial map P : k n k m.

ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES 5 (i) Given ϕ, considerp =(ϕ (Y 1 ),...,ϕ(y m )). This is a polynomial map, and we will show that P = ϕ. Letq k [Y 1,...,Y m ]: P (q(y i ) = q (P (Y i )) = q (Y i P ) = q (ϕ (Y i )) = ϕ (q (Y i )) (ii) Corollary: the set of homomorphisms k [Y 1,...,Y m ] k [X 1,...,X n ] is equivalent to the set of polynomial maps k n k m. (15) Let W 1 be an algebraic subset of k n and W 2 an algebraic subset of k m and P : k n k m a polynomial map. If P (W 1 ) W 2, then we say (what else) that P : W 1 W 2 is an algebraic map from W 1 to W 2. (a) Note that an algebraic map must be defined on the Euclidean spaces containing W 1 and W 2. (b) If we think of the polynomial functions on W i as Γ (W i ) where Γ (W 1 )= k [x 1,...,x n ] = k [X 1,...,X n ] and Γ (W 2 )=k[y 1,...,y m ] I (W 1 ) = k [Y 1,...,Y m ], I (W 2 ) then P (y i )=p i (x 1,...,x n ). (c) The map P induces a k-algebra homomorphism: P : Γ (W 2 ) Γ (W 1 ). That is, the map P defined above from all k-valued functions on W 2 to all k-valued functions on W 1 restricts to a k-algebra homomorphism Γ (W 2 ) Γ (W 1 ). (i) The proof must be obvious (ii) Proof: let F (W i ) be the k-valued functions on W i.thenk[y 1,...,y m ] is a subring of F (W 2 ) and k [x 1,...,x n ] is a subring of F (W 1 ). We know that P : F (W 2 ) F (W 1 ) is a k-algebra homomorphism (13.b.i), so it suffices to show that P sends polynomial maps to polynomial maps. For that it suffices to show that P (y i ) is a polynomial map. But P (y i )=p i (x 1,...,x n ),a polynomial map. (iii) You have to read all this very carefully and remember that all the symbols X i,x i,y i,y i are functions or maps from their respective domains to k for it to make sense. Otherwise it is just abstract nonsense or (in the words of the textbook author) rubbish. But it s not rubbish these ideas are simple, deep and important. (16) In 15 we proved that a polynomial map P : W 1 W 2 induces a homomorphism P : Γ (W 2 ) Γ (W 1 ). Nowwewanttousemorepowerful algebraic tools to prove that the set of polynomial maps is equivalent to the set of homomorphisms. That is, the assigment P Ã P is 1 1 and onto from the set of polynomial maps W 1 W 2 to the set of homomorphisms Γ (W 2 ) Γ (W 1 ). In the process of demonstrating this result we will construct the inverse equivalence. (a) Theorem: Let P : k n k m be a polynomial map, and let W 1 k n, W 2 k m be algebraic subsets. Then P (W 1 ) W 2 iff P (I (W 2 )) I (W 1 ).

6 ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES (i) Proof: P (W 1 ) W 2 P (x) W 2,allx W 1 q(p (x)) = 0, allq I (W 2 ),allx W 1 (P (q)) (x) =0,allq I (W 2 ),allx W 1 P (q) I (W 1 ),allq I(W 2 ) P (I (W 2 )) I (W 1 ) (b) Corollary: P (W 1 ) W 2 iff P lifts a homomorphism Γ (W 2 ) Γ (W 1 ). (i) Proof: 0 I (W 2 ) k [Y 1,...,Y m ] Γ (W 2 ) 0 P P ϕ 0 I (W 1 ) k [X 1,...,X n ] Γ (W 1 ) 0 By standard theorems of algebra, P induces a homomorphism ϕ iff P (I (W 2 )) I (W 1 ). (c) So every polynomial map P : W 1 W 2 induces a k-algebra homomorphism P : Γ (W 1 ) Γ (W 1 ), and every such homomorphism comes from a polynomial map. It remains to show that the assignment (polynomial maps) to (k-algebra homomorphisms) is 1 1. The choice of P in the argument above is not unique, and it is not obvious that P is unique. That is what we have to show. To do this we construct an inverse map from (k-algebra homomorphisms) to (polynomial maps) viaathirdobject. (d) If A is a finitely generated k-algebra, let MaxSpec(A) be the set of maximal ideals of A. Sincek is algebraically closed, by the Nullstellensatz there is a 1 1 correspondence between MaxSpec(Γ (W )) and W given by (a 1,...,a n ) (x 1 a 1,...,x n a n ). (i) Let W 1 k n, W 2 k m be algebraic subsets, and let ϕ : Γ (W 2 ) Γ (W 1 ) be a k-algebra homomorphism. Then ϕ induces a map MaxSpec(Γ (W 1 )) MaxSpec(Γ (W 2 )) as follows: (A) Let m MaxSpec(Γ (W 1 )). Iclaimϕ 1 (m) MaxSpec(W 2 ). because we have the induced injection: k Γ (W 2) ϕ 1 (m) Γ (W 1) m = k Therefore Γ (W 2) ϕ 1 (m) = k is a field so ϕ 1 (m) is a maximal ideal of Γ (W 2 ). Therefore the homomorphism ϕ induces amapϕ 1 : MaxSpec(Γ (W 1 )) MaxSpec(W 2 ).Next we show that from ϕ 1 we can construct an algebraic map Φ : W 1 W 2 (B) Suppose (a 1,...,a n ) W 1.Ifϕ 1 ((x 1 a 1,...,x n a n )) = (y 1 b 1,...,y m b m ) for some point (b 1,...,b m ) W 2 then ϕ (y i b i ) (x 1 a 1,...,x n a n ) or ϕ (y i )(a 1,...,a n )= b i. Thus Φ =(ϕ (y 1 ),...,ϕ(y m )) : W 1 W 2. Moreover Φ is an algebraic map. For each ϕ (y i ) k [x 1,...,x n ] lifts to a polynomial p i k [X 1,...,X n ],andp =(p 1,...,p n ):

ALGEBRAIC FUNCTIONS ON ALGEBRAIC VARITIES 7 k n k m is an algebraic map that restricts to Φ : W 1 W 2.ThemapP is not unique, but Φ is the unique algebraic map W 1 W 2 induced by ϕ. (e) We leave it to the diligent reader to verify that the correspondences between algebraic maps and homomorphisms: P Ã P ϕ Ã Φ are inverse to each other. That is, if you perform one transformation followed by the other, you get back where you started from.