[...]... proof of the theorem on factorization of integers into prime factors Irreducible components play the role of prime factors In view of Proposition 2, we can apply the previous terminology to a ne algebraic sets V Thus, we can speak about irreducible a ne algebraic k-sets, irreducible components of V and a decomposition of V into its irreducible components Notice that our topology depends very much... by means of the notion of a rational function on an a ne algebraic set First let us explain the condition that O(V ) is an integral domain We recall that V K n is a topological space with respect to the induced Zariski k-topology of K n Its closed subsets are a ne algebraic k-subsets of V From now on we denote by V (I ) the a ne algebraic k-subset of K n de ned by the ideal I k T ] If I = (F ) is... happens to be a k-algebra In particular, the eld R(V ) will be viewed as an extension k O(V ) R(V ) We will denote the eld of fractions of the polynomial ring k T1 ; : : : ; Tn ] by k(T1 ; : : : ; Tn ) It is called the eld of rational functions in n variables De nition A dominant rational k-map from an irreducible a ne algebraic k-set V to an irreducible a ne algebraic k-set W is a homomorphism of k-algebras... ) The corresponding factor-algebra k T ]=I (X ) is denoted by k X ] and is called the projective coordinate algebra of X The notion of a projective algebraic k-set is de ned similarly to the notion of an a ne algebraic k-set We x an algebraically closed extension K of k and consider subsets V Pn (K ) of the form PSol(S ; K ), where X is a system of homogeneous equations in n-variables with coe cients... on the set X (k) It assigns to a prime number p the image of m in Z=(p) = F p , i.e., the residue of m modulo p Now, we specialize the notion of a morphism of a ne algebraic varieties to de ne the notion of a regular map of a ne algebraic sets Recall that a ne algebraic k-set is a subset V of K n of the form X (K ), where X is an a ne algebraic variety over k and K is an algebraically closed extension... Prove that the correspondence K ! O(n; K ) ( = n n-matrices with entries in K satisfying M T = M ?1) is an abstract a ne algebraic k-variety 7 Give an example of a continuous map in the Zariski topology which is not a regular map 14 Irreducible algebraic sets 15 Lecture 4 IRREDUCIBLE ALGEBRAIC SETS AND RATIONAL FUNCTIONS We know that two a ne algebraic k-sets V and V 0 are isomorphic if and only if their... morphism of a ne algebraic k-varieties we have the following commutative diagram: fK ?! X (K ) = Homk (O(X ); K ) Y (K ) = Homk (O(Y ); K ) " ? ?" fO X X (O(X )) = Homk (O(X ); O(X )) ?! Y (O(X )) = Homk (O(Y ); O(X )): Take the identity map idO(X ) in the left bottom set It goes to the element in the left top set The bottom horizontal arrow sends idO(X ) to The right vertical arrow sends it to Now, because... )) = V (F ) An algebraic subsets of this form, where (F ) 6= f0g; (1), is called a hypersurface De nition A topological space V is said to be reducible if it is a union of two proper non-empty closed subsets (equivalently, there are two open disjoint proper subsets of V ) Otherwise V is said to be irreducible By de nition the empty set is irreducible An a ne algebraic k-set V is said to be reducible... must coincide with the image of under the top arrow, which is fK ( ) This proves the surjectivity The injectivity is obvious As soon as we know what is a morphism of a ne algebraic k-varieties we know how to de ne an isomorphism This will be an invertible morphism We leave to the reader to de ne the composition of morphisms and the identity morphism to be able to say what is the inverse of a morphism... the set X (K ) of K -solutions of an a ne algebraic k-variety X with homomorphisms O(X ) ! K The identi cation of this set with a subset of K n is achieved by choosing a set fo generators of the k-algebra O(X ) Forgetting about generators gives a coordinate-free de nition of the set X (K ) The correspondence K ! Hom(O(X ); K ) has the property of naturality, i.e a homomorphism of k-algebras K ! K 0