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March 30th, 2018, 08:40 AM   #1
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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.
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File Type: jpg A4ii.jpg (80.3 KB, 7 views)
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April 1st, 2018, 07:05 AM   #2
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To add on, how do I see if R/I is s field ?
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April 1st, 2018, 10:42 AM   #3
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(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.
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Last edited by cjem; April 1st, 2018 at 10:47 AM.
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April 1st, 2018, 11:18 AM   #4
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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 $I$?
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