MA3412S2_Hil2014.pdf

# The ideal p 1 is not a principal ideal of z 5 because

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The ideal P 1 is not a principal ideal of Z [ - 5] because there is no element of Z [ - 5] that is not a unit but divides both 2 and 1 + - 5. The quotient ring Z [ - 5] /P 1 is a finite field of order 2, and therefore the ideal P 1 is a maximal ideal of Z [ - 5]. The ideal P 2 1 is generated by products αβ where α and β run over a set { 2 , 1 + - 5 } of generators of the ideal P 1 . Moreover (1 + - 5) 2 = - 4 + 2 - 5. Therefore P 2 1 is the ideal of Z [ - 5] generated by 4, 2(1+ - 5) and - 4+2 - 5. Now the generators of P 2 1 are all divisible in Z [ - 5] by 2, and moreover 2 = 2(1 + - 5) - ( - 4 + 2 - 5) - 4 P 2 1 . 27

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It follows from this that P 2 1 = (2). If a , b , c and d are integers, and if a b (mod 3) then ac - 5 bd a ( c + d ) ad + bc (mod 3) . Similarly if a ≡ - b (mod 3) then ac - 5 bd a ( c - d ) ≡ - ( ad + bc ) (mod 3) . It follows that P 2 and P 3 are ideals of Z [ - 5], where P 2 = { x + y - 5 Z [ - 5] : x y (mod 3) } , P 3 = { x + y - 5 Z [ - 5] : x ≡ - y (mod 3) } . These ideals are maximal ideals of Z [ - 5], and the quotient rings Z [ - 5] /P 2 and Z [ - 5] /P 3 are finite fields of order 3. Moreover P 2 = (3 , 1 + - 5) and P 3 = (3 , 1 - - 5) . The ideal P 2 of Z [ - 5] is not a principal ideal, because there is no element of Z [ - 5] that is not a unit but that divides both 3 and 1+ - 5. Similarly the ideal P 3 is not a principal ideal. The product P 2 P 3 of the ideals P 2 and P 3 is generated by products of the form αβ , where α runs over a set { 3 , 1 + - 5 } of generators for P 2 and β runs over a set { 3 , 1 - - 5 } of generators for P 3 . Moreover (1 + - 5)(1 - - 5) = 6. It follows that P 2 P 3 is generated by 9, 3(1 + - 5), 3(1 - - 5) and 6. Therefore 3 P 2 P 3 , because 3 = 9 - 6, and thus P 2 P 3 = (3). Next we note that the ideal P 1 P 2 is generated by 6, 2(1+ - 5), 3(1+ - 5) and (1 + - 5) 2 . It follows that 1 + - 5 P 1 P 2 , because 1 + - 5 = 3(1 + - 5) - 2(1 + - 5) . Moreover 1 + - 5 divides all the listed generators of P 1 P 2 . It follows that P 1 P 2 is the principal ideal (1 + - 5) generated by 1 + - 5. Similarly P 1 P 3 = (1 - - 5). We have shown that P 2 1 = (2) and P 2 P 3 = (3). It follows that the principal ideal (6) factors as a product (6) = P 2 1 P 2 P 3 , where P 1 , P 2 and P 3 are prime ideals. This factorization of (6) as a product of prime ideals of Z [ - 5] is in fact unique. To show this, we first note that every non-zero prime ideal of Z [ - 5] is maximal. Let R = Z [ - 5], let P be a non-zero prime ideal of R , and let u + v - 5 be an element of P . Then m P , where m = N ( u + v - 5) = u 2 + 5 v 2 = ( u + v - 5)( u - v - 5) . 28
Now if a + b - 5 and c + d - 5 and if a c (mod m ) and b d (mod m ) then a + b - 5+( m ) = c + d - 5+( m ), and therefore a + b - 5+ P = c + d - 5+ P . It follows that the number of cosets of P in R cannot exceed m 2 . But a prime ideal of R that has only finitely many cosets in R must be a maximal ideal of R (Lemma 2.18). (This is a consequence of the fact that an integral domain with only finitely many elements is a field.) Thus every non-zero prime ideal of R is maximal.

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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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