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2 NEIL STRICKLANDtheorem) choose elements f 1 , . . . f m which generate it, and then V (f 1 , . . . f m ) = V (J). From nowon, we shall use only the newer notation.We can now define the Zariski topology on C n : the closed sets are precisely the sets V (J)for the various different ideals J ≤ A n . To see that this indeed defines a topology, we need thefollowing facts:∅ = V (A n )C n = V (0)V (I) ∪ V (J) = V (I ∩ J)⋂V (J k ) = V ( ∑ J k )kkAll of these are easy to prove; the third perhaps requires some comment. Suppose that a ∈V (I) ∪ V (J). Now suppose that f ∈ I ∩ J. Either a ∈ V (I), so f(a) = 0 because f ∈ I;or a ∈ V (J) and f(a)=0 because f ∈ J. Thus a ∈ V (I ∩ J). On the other hand, supposea ∉ V (I) ∪ V (J). Because a ∉ V (I), there is some f ∈ I with f(a) ≠ 0. Similarly, there is someg ∈ J with g(a) ≠ 0. Thus fg ∈ I ∩ J and fg(a) ≠ 0. Thus a ∉ V (I ∩ J). This shows thatV (I) ∪ V (J) = V (I ∩ J) as claimed.Here are some facts you can prove. I refer to the Zariski topology throughout, of course, unlessI specifically state otherwise.(1) Every one-point set {a} is closed.(2) The sets D(f) = {a | f(a) ≠ 0} form a basis for the Zariski topology.(3) D(f) ∩ D(g) = D(fg).(4) C n is not Hausdorff.(5) If we give C the Zariski topology, then the resulting product topology on C 2 is differentfrom the Zariski topology. (Consider the diagonal, for example).(6) A function f : C n −→ C m given by polynomial functions is continuous. Careful; rememberwhat you have just proved.(7) Every nonempty open set is dense.(8) Every nonempty open set is connected.Here is an interesting application of the above. Consider a 2 × 2 complex matrix( ) a bC =c dThe eigenvalues of C are the roots of the characteristic polynomialp C (λ) = det(C − λI) = λ 2 − τλ + δτ = a + dδ = ad − bcWe find that there is a repeated root iff the discriminant ∆(a, b, c, d) vanishes, where∆ = τ 2 − 4δ = (a + d) 2 − 4(ad − bc)Thus, the set V (∆) of matrices with a repeated eigenvalue is a Zariski closed subset of M 2 (C) ≃ C 4 .The complement D(∆) is open and nonempty, therefore dense. The same is true for an arbitaryvalue of 2, although some work needs to be done to define the discriminant.Now consider the function f : M 2 (C) −→ M 2 (C) defined byf(C) = p C (C) = C 2 − τC + δIThe Cayley-Hamilton theorem asserts that f(C) = 0 for any C. This is easy if C has distincteigenvalues, as we can then diagonalise everything. On the other hand, f is Zariski continuous, sof −1 {0} is closed and contains the dense set D(∆), so f −1 {0} = M 2 (C) and f = 0 as required.To return to the general theory: note that I(X) is not an entirely arbitary ideal. It has thefollowing extra property (why ?):f m ∈ I(X) ⇒ f ∈ I(X)

SOME ALGEBRAIC EXAMPLES IN TOPOLOGY 3this leads us to the following definition:√J = the radical of J = {f ∈ An | f m ∈ J for some m ∈ N}this is again an ideal. The main point in seeing this is that when expanded out, (f +g) m+l involvesonly terms divisible either by f m or by g l . Note also that√√J =√J.An ideal J is said to be a radical ideal iff J = √ J iff J = √ K for some ideal K.The full story on the correspondence between the topology and the algebra is as follows:I(V (J)) = √ JV (I(X)) = Xand the maps X ↦→ I(X) and J ↦→ V (J) give a bijection between closed sets and radical ideals.The equation I(V (J)) = √ J is essentially Hilbert’s Nullstellensatz. A “nullstelle” is a point wherea function is zero and a “satz” is a theorem. Why this has kept its German name is a mystery.There is another theorem of Hilbert in Galois theory which is for some equally mysterious reasonalways referred to as “Hilbert’s Theorem 90”. Curious.Using the above, you can show that(1) C n is compact.(2) D(f) is compact.(3) Every open subset is compact.

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