Math 610 Notes Section 1.5 1 PDF

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Download Math 610 Notes Section 1.5 1 PDF

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End of Section 1.5 then 1.6 of [Ful08],

October 4, 2017

Business • Come to the Maseeh Colloquium! – NH 454 – 3:15PM Friday, October 6, 2017 – Rafe Jones, Carleton College – Irreducibility and polynomial iteration • Conference on October 21! Oregon Number Theory Days. – Kirsten Eisentr¨ager: algebraic geometry – Graduate student: number theory and cryptography

Finishing up Section 1.5 • If V algebraic then V irreducible iff I(V ) prime. • What’s up with the radical, prime ideal I = (y 2 − x2 − (x − 1)2 )? Well we proved rad I ⊆ I(V (I) but they’re not equal in this case! • We need rad I = I(V (I) often, but that only works when fields are algebraically closed. • Algebra fact! If R is a notherian ring, then for any nonempty set I of ideals of R has a maximal member. Proof. Last time! • Who cares! • Well there’s secretly a geometry theorem we just proved! • Every collection of algebraic sets has a minimal member! • Just consider {I (V ) ∣ V in your collection}; it has a maximal element. • Theorem: Every algebraic set can be decomposed uniquely into irreducibles. Ie, there is a unique set of irreducible algebraic sets {F1 , . . . , Vm } such that for all i ≠ j in {1, . . . , m}, we have that Vi ⊈ Vj .

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Proof.

– Let S = {sets that can’t be written as a finite union of algebraic sets} .

– So S has no irreducible algebraic sets! – Suppose for a contradiction that S ≠ ∅. – The we know that S has a minimal element, let’s call it V . – Not irreducible, so write it as V = V1 ∪ V2 , with Vi ⊆ V . – But then the Vi aren’t in S because V was minimal! – so we can write them as a finite union. – But then V is a finite union, which is our contradiction, because V ∈ S . – OK, how do we know no sets contain each other? Just throw some away! – Uniqueness. – Suppose that V = V 1 ∪ ⋯ ∪ V m1 = W 1 ∪ ⋯ ∪ W m2 Well, V1 = ∪i (V1 ∩ Wi ), so then there’s an i with V1 = Wi . – And there’s a j with Wi = Vj . – But this means that Vj = V1 = Wi , so j = 1 and get rid of the V1 and Wi and continue in this way.

• V (y 4 − x2 , y 4 − x2 y 2 + xy 2 − x3 )

1.6 Algebraic subsets of the plane Let’s focus on the case A2 (k) for a while. Proposition 1. Suppose that f, g ∈ k[x, y] and f, g have no common factors. Then V (f, g) = V (f ) ⋂ V (g) is a finite set of points. Proof. Recall a version of Gauss’s lemma: Suppose that R is a UFD and f, h ∈ R[x] and h primitive. If h ∣ f in Frac (R)[x], then h ∣ f in R[x]. It has a corollary: if f, g have no common factor in R[x], then they have no common factor in Frac (R)[x]. Well, first note that f, g have no common factor in k[x][y ], so they have no common factor in k(x)[y ], which is a Euclidean domain! (Mention how we use the “factor” theorem in multivariate polynomials; it’s by thinking about multivariate polynomials as univariate polynomials over rational function fields.) Thus, there are some r, s ∈ k(x)[y] such that rf + sg + 1. Canceling denominators, there is some d ∈ k[x] such that dr, ds ∈ k[x, y], so then (dr )f + (ds)g = d.

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Suppose that (a, b) ∈ V (f, g ), so then 0 = (dr)f (a, b) + (ds)g (a, b) = d(a, b) = d(a). But there are only a finite number of zeros of d, so there are only a finite number of options for a. The same reasoning applies to y-coordinates, which completes the proof. Corollary 2. Suppose that f ∈ k[x, y] is irreducible and V (f ) is infinite. Then I(V (f )) = (f ) and V (f ) is irreducible Proof. For the first statement, clearly (f ) ⊆ I (V (f )), so suppose that g ∈ I(V (f )). Then V (f, g) is infinite, so f, g have a common factor by Proposition 1. But f is irreducible, so that factor must be f , so g ∈ (f ). The fact that V (f ) is irreducible follows from the fact that (f ) is prime. (Good thing that k[x, y] is a UFD.) Corollary 3. Suppose that k is infinite. Then the irreducible algebraic subsets of A2 (k) are • A 2 (k ). • ∅, • points, and • irreducible plane curves; i.e., algebraic sets of the form V (f ) for f ∈ k[x, y] irreducible with V (f ) infinite. Proof. Suppose that V is an irreducible algebraic subset of A2 (k ), and V isn’t A2 (k ), or a point, or ∅ (all of which are clearly irreducible). Well then, I(V ) ≠ (0), so I(V ) contains a nonconstant polynomial; let’s call it f . Since I (V ) is prime, one of the irreducible factors of f is in I , so we can assume that f is irreducible. I claim that I(V ) = (f ). To see this, suppose that g ∈ I(V ). If g ∉ (f ), then f, g have no common factor, so V (f, g) is finite by Proposition 1. But V (f, g) ⊇ V , so V is finite. But that means that V is a point or the empty set, which we assumed that it was not. (Algebraic sets consisting of many points are clearly irreducible.) Corollary 4. Suppose that k is algebraically closed, and choose some non constant f ∈ k[x, y]. Let’s say that f factors into irreducibles as f = f1a1 ⋯fnan . Then the decomposition of V (f ) into irreducible components is V (f ) = V (f1 ) ⋃ ⋯ ⋃ V (fn ) and I(V (f )) = (f1 ⋯fn ) . 3

Proof. Recall that each V (fi ) is infinite since k is algebraically closed (this is exercise 1.14, guiding exercise). Compare to the example at start. Well then, all the V (fi )s are irreducible since all the fi s are irreducible (Corollary 2). Moreover, no V (fi ) includes any other since no ideal contains any other, so the first statement is proved. By Corollary 2, then, I (V (fi )) = (Fi ), so I(V (f )) = I (V (f1 ) ⋃ ⋯ ⋃ V (fn )) = I (V (f1 )) ⋂ ⋯ ⋂ I (V (fn )) = (f1 ) ⋂ ⋯ ⋂ (fn ) by . Finally, if you’re divisible by each fi , then you’re divisible by f1 ⋯fn , so we’re done! (It’s not generally true that I ⋂ J = I J, but lucky us we’re in a UFD.)

References [Ful08] William Fulton, Algebraic curves: An introduction to algebraic geometry, 2008.

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