MA3412S2_Hil2014.pdf

# Proof let i be a non zero proper ideal of r then

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Proof Let I be a non-zero proper ideal of R . Then there exists some non- zero element x of R such that I = ( x ). Moreover x is not a unit of R . The principal ideal domain R is a unique factorization domain (Lemma 2.23). Therefore there exist prime elements p 1 , p 2 , . . . , p k such that x = p 1 p 2 · · · p k . Let P i = ( p i ) for i = 1 , 2 , . . . , k . Then each ideal P i is a non-zero prime ideal of R , and I = P 1 P 2 · · · P k . Two prime elements of R generate the same prime ideal of R if and only if they are associates. The uniqueness of the factorization of I as a product of prime ideals therefore follows directly from Proposition 2.21. 2.8 An Integral Domain lacking Unique Factorization The integral domain Z [ - 5] consists of all all complex numbers that are of the form x + y - 5 for some integers x and y . We define the norm N ( x + y - 5) of an element x + y - 5 of Z [ - 5] to be x 2 + 5 y 2 . The norm N ( ω ) of an element ω of Z [ - 5] is thus a non-negative integer. Moreover N (( x + y - 5)( u + v - 5)) = N ( xu - 5 yv + ( xv + yu ) - 5) = ( xu - 5 yv ) 2 + 5( xv + yu ) 2 = x 2 u 2 + 25 y 2 v 2 + 5 x 2 v 2 + 5 y 2 u 2 = ( x 2 + 5 y 2 )( u 2 + 5 v 2 ) = N ( x + y - 5) N ( u + v - 5) for all integers x , y , u and v . Thus N ( ωθ ) = N ( ω ) N ( θ ) for all ω, θ Z [ - 5]. If ω is a unit of the integral domain Z [ - 5] then ω - 1 Z [ - 5], and therefore N ( ω ) and N ( ω - 1 ) are both positive integers satisfying N ( ω ) N ( ω - 1 ) = N (1) = 1 . 26

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It follows that if ω is a unit of Z [ - 5] then N ( ω ) = 1. From this it follows that the only units of Z [ - 5] are 1 and - 1. Now 6 = 2 × 3 = (1 + - 5)(1 - - 5) , where the factors 2, 3 and 1 ± - 5 are all elements of Z [ - 5]. Now N (2) = 4. Now there is no element ω of Z [ - 5] that satisfies N ( ω ) = 2 or N ( ω ) = 3. Thus if ω and θ are elements of Z [ - 5] satisfying ωθ = 2 then N ( ω ) N ( θ ) = 4, and therefore either N ( ω ) = 1, in which case ω is a unit of Z [ - 5], or else N ( θ ) = 1, in which case θ is a unit of Z [ - 5]. It follows that the integer 2 is an irreducible element of Z [ - 5]. Analogous arguments show that 3, 1 + - 5 and 1 - - 5 are irreducible elements of Z [ - 5]. The irreducible elements 2, 3, 1 + - 5 and 1 - - 5 of Z [ - 5] are not prime elements of Z [ - 5]. Indeed 2 and 3 divide the product of 1+ - 5 and 1 - - 5, but do not divide either factor of this product. Similarly 1 + - 5 and 1 - - 5 divide the product of 2 and 3 but do not divide either 2 or 3. The principal ideals generated by these elements are neither prime nor maximal. Let a , b , c and d be integers. Then ( a + b - 5)( c + d - 5) = ac - 5 bd + ( ad + bc ) - 5 . If a b (mod 2) then ac - 5 bd a ( c + d ) (mod 2) and ad + bc a ( c + d ) (mod 2). It follows that ac - 5 bd ad + bc (mod 2) whenever a b (mod 2). Thus if P 1 = { x + y - 5 Z [ - 5] : x y (mod 2) } then P 1 is an ideal of Z [ - 5]. This ideal contains the elements 2, 1+ - 5 and 1 - - 5. Given any element x + y - 5 of P 1 there exists u + v - 5 Z [ - 5] such that x + y - 5 - 2( u + v - 5) ∈ { 0 , 1 + - 5 } . It follows that P 1 = (2 , 1 + - 5) = (2 , 1 - - 5) .
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