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March 31st, 2018, 07:07 AM   #1
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Joined: Sep 2011

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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.
Attached Images Q2...jpg (21.6 KB, 6 views) Q2i.jpg (83.2 KB, 6 views) Q2ii.jpg (64.1 KB, 5 views) Q2iii.jpg (92.1 KB, 4 views) March 31st, 2018, 09:25 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic 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) = $. 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 non-zero elements are invertible: when you say "suppose $a, b \neq 0$, did you mean "suppose $a$ and $b$ are both non-zero" or "suppose $a$ or $b$ is non-zero? 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 $a + 0 \sqrt{2}$ ($a$ non-zero, rational) nor $0 + b \sqrt{2}$ ($b$ non-zero, 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. Thanks from Alexis87, topsquark and Country Boy Tags congruence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gerva Number Theory 6 August 11th, 2016 11:26 PM s.a.sajib Number Theory 9 April 8th, 2013 06:28 AM justjones Algebra 2 March 13th, 2013 02:38 PM Fernando Number Theory 3 May 15th, 2012 08:20 AM Aizen Number Theory 0 June 28th, 2008 12:05 PM

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