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Math Help - Coprime Question

  1. #1
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    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?
    Last edited by gummy_ratz; March 26th 2011 at 08:41 AM.
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  2. #2
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    Quote Originally Posted by gummy_ratz View Post
    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 ij\,,\,i\in I\,,\,j\in J.

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

    Tonio
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    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
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  4. #4
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    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!
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    "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.
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  6. #6
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    Right. You have to define IJ:=\{\sum_{\mathrm{finite}}i_kj_k | i_k\in I,j_k\in J\} if you want IJ to be an ideal. The set \{ij | i\in I,j\in J\} will not be an ideal in general, as it will not be closed under sums.
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