March 31st, 2018, 07:07 AM  #1 
Member Joined: Sep 2011 Posts: 97 Thanks: 1  Congruence in F[x]
Hi all, I have completed this question as attached. Hope someone could help to check if my solutions are correct. However, I am not sure what theorem is used for part (d). May need some advice. Thanks. Note: Please click on the thumbnails to enlarge. 
March 31st, 2018, 09:25 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 204 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  (a) Injectivity. You should justify the implication $[a + bx] = [c + dx] \implies a = c$ and $b = d$. Note that $[g(x)] = [h(x)]$ doesn't mean $g(x) = h(x)$, but that $g(x)$ and $h(x)$ differ by a multiple of $x^2  2$. Also, to prove a homomorphism $f$ of rings (or of groups/algebras/modules/etc.) is injective, it suffices to show its kernel is trivial (i.e. that $f(\alpha) = 0$ implies $\alpha = 0$). Surjectivity. I think you've got the right idea, but it looks like you've got a "typo". You've suggested $f(a + b \sqrt{2}) = A$ is an element of $\mathbb{Q}(\sqrt{2})$ rather than of $R$. Homomorphism. Depending on your definition of a ring homomorphism, you might want to show $f(1) = [1]$. This seems fine other than that. (b) You're given that $\mathbb{Q}(\sqrt{2})$ is a commutative ring with unity, so showing closure under addition and multiplication is unnecessary. One small thing with showing nonzero elements are invertible: when you say "suppose $a, b \neq 0$, did you mean "suppose $a$ and $b$ are both nonzero" or "suppose $a$ or $b$ is nonzero? If you meant the latter, your argument is fine, but you should probably phrase it a bit more clearly. If you meant the former, then you haven't shown that the elements $a + 0 \sqrt{2}$ ($a$ nonzero, rational) nor $0 + b \sqrt{2}$ ($b$ nonzero, rational) are invertible. An alternative way to see that it's a field is to note that $(x^2 2)$ is a maximal ideal of $\mathbb{Q[x]}$ so $R$ is a field. (c) This looks fine. (d) You should be careful here. Just because the inverse of $1 + 2 \sqrt{2}$ has a certain form, you can't just assume that the inverse of $[1+2x]$ should have the same form. A way to prove it is to use the fact that, if $ \phi: S \to T$ is an isomorphism of rings and $u \in S$ is invertible, then $\phi(u)$ is invertible in $T$ and $\phi(u)^{1} = \phi(u^{1})$. Indeed, this gives $[1+2x]^{1} = f(1 + 2 \sqrt{2})^{1} = f((1+2 \sqrt{2})^{1}) = f \left( \frac{1}{7} + \frac{2}{7} \sqrt{2} \right) = \left[ \frac{1}{7} + \frac{2}{7} x \right]$ _________________ I hope this helps. If anything is unclear, please let me know and I'll try my best to clarify. 

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