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Math Help - Rings and domains

  1. #1
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    Rings and domains

    Hi guys. Struggling with a few bits and pieces. Can't seem to get these from my notes or text book. If someone could talk me through them it would be very much appreciated

    [IMG]file:///C:/DOCUME%7E1/Jasdeep/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Jasdeep/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]35. Show that for D = 3, 6, 7 the group of units (Z[√D]) in the ring Z[√D] is
    infinite by exhibiting infinitely many units in each of the rings.

    38. Let  : Q[X] → Q(√3) be the map defined by (a0 + a1X + ... + anXn) =
    (a0 + a1√3 + ... + an(√3)n). Show that  is a surjective ring homomorphism. Prove that Ker  = (X2 − 3)Q[X]. Deduce that the factor ring Q[X]/Ker  is isomorphic to Q(√3).

    40. Let  : R → S be a ring homomorphism. Prove that if u ∈ R is a unit then
    (u) ∈ S is a unit and that (u−1) = (u)−1.

    44. Show that the ideal (2, 1+√−5)R of the ring R = Z[√−5] can not be principal.

    47. Prove that Z[√D] is an Euclidean domain, with respect to the norm defined by
    N(a + ib) = a2 − Db2 in the following cases D = −2,−3, 2, 3, 6, 7.
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  2. #2
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    Quote Originally Posted by jdizzle1985 View Post
    38. Let  : Q[X] → Q(√3) be the map defined by (a0 + a1X + ... + anXn) =
    (a0 + a1√3 + ... + an(√3)n). Show that  is a surjective ring homomorphism. Prove that Ker  = (X2 − 3)Q[X]. Deduce that the factor ring Q[X]/Ker  is isomorphic to Q(√3).
    You need to prove that I(f(x)g(x)) = I(f(x))I(g(x)) this follows because of how we defined polynomial multiplication. It is surjective because if \alpha \in \mathbb{Q}(\sqrt{3}) then it means \alpha = a+b\sqrt{3} and so I(a+bx) = x. The kernel are all f(x) so that f(\sqrt{3}) = 0, since x^2 - 3 is an irreducible polynomial it must divide any f(x) with f(\sqrt{3})=0 this means \left< x^2 - 3\right> = \ker I. Now by fundamental homomorphism thereom for rings the results follows.
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    Quote Originally Posted by jdizzle1985 View Post
    40. Let  : R → S be a ring homomorphism. Prove that if u ∈ R is a unit then
    (u) ∈ S is a unit and that (u−1) = (u)−1.
    Let \phi: R\mapsto S be a ring homomorphism. Let u be a unit in R then it means there exists v\in R so that uv=1 but then \phi (uv) = \phi(1)\implies \phi(u)\phi(v) = 1' and so \phi(v) is the inverse for \phi(u). Furthermore this shows \phi(u^{-1}) = (\phi(u))^{-1}.
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