1. ## 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.

2. Originally Posted by jdizzle1985
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.

3. Originally Posted by jdizzle1985
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 $\phi(u^{-1}) = (\phi(u))^{-1}$.