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March 30th, 2018, 08:40 AM   #1
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Joined: Sep 2011

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Ideals Problem

I have attached my solution for part (a), (b) and (c). I am not sure if part (a), (b) are correct. However for part (c), the question did not define the output of the function so I am not sure if I can do it as such. Therefore need verification on all the 3 parts. Thanks! Please click on the thumbnails to enlarge.
Attached Images A4Q4.jpg (21.3 KB, 7 views) A4ii.jpg (80.3 KB, 7 views) A4i.jpg (77.5 KB, 4 views) April 1st, 2018, 07:05 AM #2 Member   Joined: Sep 2011 Posts: 99 Thanks: 1 To add on, how do I see if R/I is s field ? April 1st, 2018, 10:42 AM #3 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry (a) This is fine. Note that it's not actually necessary to show condition (iii) as this is automatically implied by (i) and (ii). To be completely clear, what I'm saying is: Suppose $S$ is a ring and $A$ is a non-empty subset such that: (i) $a,b \in A \implies a+b \in A$ (ii) $a \in A, s \in S \implies sa \in A$ Then $a \in A \implies -a \in A$. If this isn't immediately obvious to you, try to prove it. (b) It seems like you've got the idea, but you should be careful with writing out your arguments. For example, you've written "Then $\begin{pmatrix} k & p \\ 0 & k \end{pmatrix} \equiv \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \mod I$ $\begin{pmatrix} k & p \\ 0 & k \end{pmatrix} - \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = \begin{pmatrix} 0 & p \\0 & 0 \end{pmatrix} \in I$". It would be good if you indicated that the second line implies the first. You could simply add a word like "because" at the start of the second line. Also, you've said "$R/I$ can be written in the form..." where you probably mean "The elements of $R/I$ can be written in the form...". (c) The question didn't define the output of the function because it's asking you to do that! Start by doing that, and then see if you can finish the problem from there. Also, at the top of the page, it looks like you're treating $f$ as a function from $\mathbb{Z} \times \mathbb{Z}$ to $R$ rather than from $R$ to $\mathbb{Z}$; be careful! (d) Once you've shown that $R/I$ and $\mathbb{Z}$ are isomorphic rings, the question simply becomes: is $\mathbb{Z}$ is a field? Hopefully you know the answer to this! Edit: To follow what I've said in part (d), you would need to know: if $A$ and $B$ are isomorphic rings, then $A$ is a field if and only of $B$ is a field. Hopefully you either already know this or can at least prove it without difficulty. Thanks from Alexis87 Last edited by cjem; April 1st, 2018 at 10:47 AM. April 1st, 2018, 11:18 AM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics For part (c) you haven't defined a homomorphism so I don't see how you can decide it is surjective. Try the mapping $\left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right) \mapsto a.$ Do you see why this is the obvious choice? Can you prove this is a homomorphism whose kernel is exactly $I$? Thanks from Alexis87 Tags ideals, problem 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 shashank dwivedi Abstract Algebra 1 December 22nd, 2017 01:58 AM kp100591 Abstract Algebra 1 May 19th, 2014 06:09 AM Lolyta Abstract Algebra 1 September 30th, 2013 04:35 AM chappyform Abstract Algebra 1 March 11th, 2013 05:05 AM tiger4 Abstract Algebra 1 April 21st, 2012 09:27 AM

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