An example of primary decomposition Let k be a field and R = k[x; y; z], the polynomial ring in three variables. Let I ⊂ R be the ideal given by I = (x2z2; x(x + y2); z(z − y2)) a) A minimal primary decomposition of I is given by I = (x; z) \ (x; z − y2) \ (z; x + y2) \ (y4; x(x + y2); z(z − y2)) where the three first components are prime ideals, and the last one is (x; y; z)- primary, i.e. p(y4; x(x + y2); z(z − y2)) = (x; y; z). Consequently Ass(R=I)) = f(x; z); (x; z − y2); (z; x + y2); (x; y; z)g where the three first are minimal, and the last one is embedded. Proof. We have x2z2 = (x2 + xy2 − xy2)(z2 − zy2 + zy2) = (x(x + y2) − xy2)(z(z − y2) + zy2) = x(x + y2)z(z − y2) + x(x + y2)zy2 − xy2z(z − y2) − xzy4 and it follows that I ⊂ (y4; x(x+y2); z(z−y2)). Obviously I ⊂ (x; z), I ⊂ (x; z−y2) and I ⊂ (z; x + y2). Thus I ⊂ (x; z) \ (x; z − y2) \ (z; x + y2) \ (y4; x(x + y2); z(z − y2)) 4 2 2 2 2 On the other hand, if F = F1y +F2x(x+y )+F3z(z −y ) 2 (x; z−y )\(z; x+y ), then 2 2 2 2 F1 = G1xz + G2x(x + y ) + G3z(z − y ) + G4(x + y )(z − y ) It follows that if F 2 (x; z), then G4 2 (x; z), i.e. G4 = H1x + H2z. Putting all together we get 4 4 2 4 2 4 y 4 2 F = G1xzy +[G2y +F2+H1(z−y )y ]x(x+y )+[G3y +F3+H2(x+y )y ]z(z−y ) which shows that F 2 I. The radical p(y4; x(x + y2); z(z − y2)) = p(y; x2 + xy2; z2 − zy2) = p(y; x2; z2) = (x; y; z) It is easily seen that the decomposition is minimal, as the four primes are distinct, and removing any of them will change the ideal. b) By the first uniqueness theorem we know that Ass(R=I) consists of the prime ideals in R of the form Ann(fi) in R=I for elements fi 2 R. In this example we have the following: (x; z) = Ann((x + y2)(z − y2)) (x + y2; z) = Ann(x(z − y2)) (x; z − y2) = Ann((x + y2)z) (x; y; z) = Ann(αx2yz + βxyz2) α; β 2 k 1 2 Proof. The three first are more or less obvious. For the last one it is easily seen that x2yz; xyz2 62 I, since there is no monomial with only one y in the ideal. But x · x2yz = x(x2 + xy2 − xy2)yz = x(x(x + y2))yz − x2z2y3 2 I z · x2yz = x2z2y 2 I y · x2yz = x2y2z = x2(y2 − z + z)z = x2(y2 − z)z + x2z2 2 I c) By the second uniqueness theorem we can also compute the primary compo- nents of the minimal primes, as given in the decomposition. The components are given by the formula S Qi = I = fF 2 k[x; y; z] j 9s 2 S s:t: sF 2 Ig where S = R n pi. In fact we have for (x; z) that x · (x + y2) 2 I; and x + y2 62 (x; z) and z · (z − y2) 2 I; and z − y2 62 (x; z) Similar for the two other minimal prime ideals. d) Suppose k is algebraically closed. Then I ⊂ m = (x − a; y − b; z − c) if and only if the triple (a; b; c) 2 k3 satisfies one of the following equations: i) a = c = 0 i) a = 0, c = b2 i) c = 0, a = −b2 Proof. If I = (x2z2; x(x + y2); z(z − y2)) ⊂ (x − a; y − b; z − c), then φ(I) = 0 where φ : R=I ! R=m =∼ k is given by x 7! a, y 7! b and z 7! c, i.e. a2c2 = a(a + b2) = c(c − b2) = 0 2 2 If c 6= 0, then c = b and a = 0, and if a 6= 0, then a = −b and c = 0. e) The set Supp(R=I) is the union of Ass(R=I) and all the maximal ideals in d). Proof. We know that for the finitely generated R-module R=I, Supp(R=I) = V (Ann(R=I)) But Ann(R=I) = I, and Supp(R=I) equals the set of all prime idealsp which contains I. Let p ⊂ R be a prime ideal which contains I. Then we have I ⊂ p since a prime ideal is its own radical. Thus we have p I = (x; z) \ (x; z − y2) \ (z; x + y2) \ (x; y; z) ⊂ p Tm In general we know that if i=1 ai ⊂ p, then ai ⊂ p for some i = 1; : : : ; m. The argument goes as follows: Suppose ai 6⊂ p for all i. Then for each i there exist Q Q T T ai 2 ai, but ai 62 p. Thus a = ai 2 ai ⊂ ai, but a 62 p. Hence ai 6⊂ p. For the three non-maximal prime ideals pi the quotient R=pi is isomorphic to a polynomial ring in one variable. In k[t] every non-trivial prime ideal is maximal, thus p has to be maximal, i.e. in the set Supp(R=I). .