1. ## Coprime Question

On page 67 here -> http://www1.spms.ntu.edu.sg/~frederique/AA10.pdf
I understand why InJ is a subset of IJ right up to ax + ay is an element of IJ.
I figured that ax + ay = yx + xy (since a is in I and a is in J) so we have xy and yx are in IJ... but how do we know that when you add two elements of IJ, you still get an element of IJ... like, how do we know xy + xy is an element of IJ?

2. Originally Posted by gummy_ratz
On page 62 here -> http://www1.spms.ntu.edu.sg/~frederique/AA10.pdf
I understand why InJ is a subset of IJ right up to ax + ay is an element of IJ.
I figured that ax + ay = yx + xy (since a is in I and a is in J) so we have xy and yx are in IJ... but how do we know that when you add two elements of IJ, you still get an element of IJ... like, how do we know xy + xy is an element of IJ?

Definition: the elements of IJ are all the finite sums of elements of the form $\displaystyle ij\,,\,i\in I\,,\,j\in J$.

You seem to believe that the elements in IJ are uniquely of the form $\displaystyle ij$ , and this is wrong.

Tonio

3. The first thing I would like to do is thank gummy_ratz for the notes. I'm currently studying this and the extra reference will be valuable. (Can't have too many textbooks! )

I'm sure tonio is correct in his analysis, however I have looked at page 62 and I have to admit defeat. What material on page 62 refers to the question??

-Dan

4. Oo I'm sorry, I meant page 67! My bad. I'll edit my post. Ohh okay, that proof makes sense with that definition of IJ. Thanks!

5. "IJ contains by definition sums of elements xy, x in I, y in J , with xy in I and xy in J by definition of two-sided ideal."

this is from page 67.

6. Right. You have to define $\displaystyle IJ:=\{\sum_{\mathrm{finite}}i_kj_k | i_k\in I,j_k\in J\}$ if you want $\displaystyle IJ$ to be an ideal. The set $\displaystyle \{ij | i\in I,j\in J\}$ will not be an ideal in general, as it will not be closed under sums.