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5. Rings May 5, 2008 1 / 12 Ideals: 1847, FLT, Etc

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5. Rings May 5, 2008 1 / 12 Ideals: 1847, FLT, Etc Ideals: Definitions & Examples Defn: An ideal I of a commutative ring R is a subset of R such that for a, b ∈ I and r ∈ R we have a + b, a − b, ra ∈ I Examples: All ideals of Z have form nZ = (n) = {..., −n, 0, n, 2n,...} All ideals of [i] have form α [i] = (α) = {αβ|β ∈ [i]} for α ∈ [i] √ Z Z Z Z [ −5] has ideals of form: Z √ √ √ αZ[ −5] = (α) = {αβ|β ∈ Z[ −5]}√for α ∈ Z[ −5] √ (α1, α2) = {α1β1 + α2β2|β1, β2 ∈ Z[ −5]} for α1, α2 ∈ Z[ −5] Z[x] has the ideal (2, x) - all polynomials with even constant terms. Robert Campbell (UMBC) 5. Rings May 5, 2008 1 / 12 Ideals: 1847, FLT, etc Lam´e’s“Proof”: In 1847 Lam´epresented a proof of Fermat’s Last Theorem with a rough outline of: Given ap + bp = cp 2 p−1 p Factor as (a − b)(a − ξpb)(a − ξpb) ... (a − ξp b) = c where 2π 2π ξp = cos( p ) + i sin( p ) is a primitive pth root of unity and Z[ξp] is the Cyclotomic integers. If {(a − ξk b)} are coprime then each of the (a − ξk b) must be a pth power. If they are not coprime, divide by the common factor and repeat (infinite descent). Gap: Proof assumes that if x, y are irreducible and xy = zn then x, y are nth powers - depends on unique factorization, but Z[ξ23] does not have unique factorization. Solution: Ideal numbers. Robert Campbell (UMBC) 5. Rings May 5, 2008 2 / 12 Kummer & Dedekind Kummer (1847) proposed that unique factorization could be recovered by adding new elements to the ring - “ideal numbers”. Dedekind published (1876) the concept of ideal as a set of elements preserved under addition, negation and multiplication, which could be thought of as the set of multiples of an ideal number. Hilbert’s Example: Consider the set S = {n ≡ 1(mod 4)} = {1, 5, 9,...} Closed under multiplication (ignore addition) Irreducibles are P = {5, 9, 13, 17, 21, 29,..., 49,...} 441 = (21)(21) = (9)(49) 2 2 [over Z gcd(9, 21) = 3, gcd(21, 49) = 7, so 441 = (3 )(7 )] Kummer Ideal Number: Add 3 and 7 to the set (multiples?) Dedekind Ideal: Replace numbers with sets: [3] = {9, 21, 33, 45,...} [7] = {21, 49, 77,...} Robert Campbell (UMBC) 5. Rings May 5, 2008 3 / 12 Ideals & Ideal Numbers: An Example √ Example: Consider√Z[ −5] √ 6 = (2)(3) = (1 + −5)(1 − −5) √ √ Ideals: Factor√ 3, 2, (1√ + −5) and (1 − −5) as ideals (3) = (3, √−5 + 1)(3, −5 − 1) 2 (2) =√ (2, −5 +√ 1) √ (1 + √−5) = (2, √−5 + 1)(3, √−5 + 1) (1 − −5) = (2, −5 + 1)(3, −5 − 1) √ Ideals: adjoin 2 to the ring Robert Campbell (UMBC) 5. Rings May 5, 2008 4 / 12 Ideals: Generating Sets Defn: The ideal of R generated by set X is the intersection of all ideals of R containing X , I (X ) = ∩{I |X ⊆ I , I ideal ⊆ R}. (also denoted (X )) PN Prop: For X finite, I (X ) = { i=0 βi xi |βi ∈ R}. pf: I (X ) is closed under addition and multiplication by R, so it is an ideal. X ⊆ I =⇒ I (X ) ⊆ I as I must be closed. Robert Campbell (UMBC) 5. Rings May 5, 2008 5 / 12 Principal Ideals Defn: An ideal is principal if it has some generating set of one element, A = (α) Examples: All ideals of Z have form nZ = (n) and are principal Which can also be written as (n) = (n, 3n, 5n) = (3n, 7n) All ideals of [i] have form α [i] = (α) and are principal √ Z Z [ −5] has ideals of form: Z √ √ √ αZ[ −5] = (α) = {αβ|β ∈ Z[ −5]}√for α ∈ Z[ −5] (principal)√ (α1, α2) = {α1β1 + α2β2|β1, β2 ∈ Z[ −5]} for α1, α2 ∈ Z[ −5] (non-principal) Z[x] has the ideal (2, x) - all polynomials with even constant terms. (non-principal) Robert Campbell (UMBC) 5. Rings May 5, 2008 6 / 12 PIDs Defn: A principal ideal domain (PID) is an integral domain, all of whose ideals are principal. Examples: Z is a PID F[x] is a PID Z[x] is not a PID (2, x) ⊂ Z[x] is not principal Robert Campbell (UMBC) 5. Rings May 5, 2008 7 / 12 Algebra of Ideals Defn: Robert Campbell (UMBC) 5. Rings May 5, 2008 8 / 12 Prime & Maximal Ideals Defn: An ideal of R is prime if for any a, b ∈ R, if ab ∈ I , then either a ∈ I or b ∈ I Defn: An ideal I of R is maximal if I ⊂ J for some ideal J implies J = I or J = R Prop: If I is a maximal ideal of R then I is a prime ideal of R. Prop: If I is a non-zero prime ideal of PID R then I is a maximal ideal of R. Examples: Z:(n) ⊂ Z is maximal and prime iff n ∈ Z is prime [x]: (f (x)) ⊂ [x] is maximal and prime iff f (x) ∈ [x] is irreducible Q √ √ q √ Q Q[ 6]: (2, 6) ⊂ Q[ 6] is prime Robert Campbell (UMBC) 5. Rings May 5, 2008 9 / 12 Polynomial Ideals n Defn: An affine algebraic set S is the set of points in the vector space F which satisfy a finite set of polynomial equations {fi (x0,..., xn) = 0}. Defn: An algebraic set S is reducible iff it can be expressed as the union of two proper subsets, each of which is an algebraic set, S = S1 ∪ S2 Defn: An algebraic variety is an irreducible algebraic subset. Examples: The union of curves {(x, y)|y = x2 + 1} ∪ {(x, y)|y = 2x} is an algebraic set but not a variety Each curve, {(x, y)|y = x2 + 1} or {(x, y)|y = 2x} is an algebraic variety Prop: The set of polynomials in F[x0,..., xn] which is satisfied by all points in some variety V is an ideal, I (V ). Prop: An algebraic variety V is irreducible iff I (V ) is a prime ideal. Robert Campbell (UMBC) 5. Rings May 5, 2008 10 / 12 Examples: The centered circle of radius 2 in the plane z = 2 satisfies: {x2 + y 2 − 4 = 0, z − 2 = 0} (cylinder & plane) {x2 + y 2 + z8 − 8 = 0, z − 2 = 0} (sphere & plane) {x2 + y 2 − 4 = 0, x2 + y 2 − 2z = 0} (cylinder & paraboloid) {x2 + y 2 − 4 = 0, x2 + y 2 + z8 − 8 = 0, x2 + y 2 − 2z = 0} Quotient Construction Lemma: The relation r1 ∼ r2 iff r1 + I = r2 + I is an equivalence relation. Defn: The quotient ring is R/I = {r + I }/ ∼, where addition and multiplication is inherited from R. Prop: R/I is a ring Examples: Zn = Z/nZ Robert Campbell (UMBC) 5. Rings May 5, 2008 11 / 12 Quotients: Prime & Maximal Ideals Prop: R/P is an integral domain iff P ⊂ R is a prime ideal Prop: R/M is a field iff M ⊂ R is a maximal ideal Robert Campbell (UMBC) 5. Rings May 5, 2008 12 / 12.
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