Question

Abstract Algebra problem

Answer 1

I concluded that r and t must be even, otherwise multiplying an element of S with any element of R does not necessarily produce another element in S. Does that solution sound reasonable?

Answer 2

The problem is attached as a picture.

Answer 3

For some reason, when I saw the word 'algebra', I thought it'd be something fun and simple, like x and y, ya know :D

Answer 4

The title "abstract algebra" is pretty misleading. I think most people assume we are just working more with algebra from high school when we say we are taking abstract algebra. It's quite a different world.

Answer 5

<snaps fingers>
yes yes yes
abstract algebra is even MORE fun ^.^

Now, S is an ideal, for any element X in R and Y in S, we can be sure XY is an element of S, right? ^.^

Answer 6

As well as YX, according to how our text defines ideals.

Answer 7

Okay, let's see where that takes us (I still don't know. I accidentally ignored your first post on purpose, so that I have no preconceptions)
s is even, so I might as well let s = 2k, for some integer k, for whatever good it does me.

We have
\[\Large \left[\begin{matrix}a & b \\ 0 & d\end{matrix}\right]\left[\begin{matrix}r & 2k \\ 0 & t\end{matrix}\right]\in S\]

Answer 8

\[\Large \left[\begin{matrix} ar & 2ak+tb \\ 0 & td\end{matrix}\right]\in S\]\

Answer 9

means tb has to be even, regardless of b.
Okay, so t has to be even too, I suppose, otherwise, it won't work if b were odd :/

Answer 10

Right. And when I multiplied in the reverse order, I got rb + sd as the entry in row 1 column 2. sd is even regardless of b, but r must be even to ensure that the entire entry is always even and thus an element of S. Right?

Answer 11

Makes sense.
And that was your first post.

So... what did you need help with again? :D

Answer 12

When I posted the first time, I second-guessed my solution and decided to work through the problem again to make sure I didn't miss anything. The solution seemed to little simple, so I was skeptical. I tend to be skeptical of solutions that seem too easy. Anyways, thanks for your help. I like to get second opinions on problems in this class.

Answer 13

yay abstract algebra :3

Well, not really

I'm so confused.

Hi ^.^