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Math Help - Ideals in Q[x,y]

  1. #1
    Super Member Bernhard's Avatar
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    Ideals in Q[x,y]

    I am reading Chapter 1, Section 4, Cox et al "Ideals, Varieties and Algorithms"

    Exercise 3(a) reads as follows:

    In  \mathbb{Q}[x,y] show the following equality of ideals:

    <x + y, x - y > = <x, y>

    I would appreciate help with this problem.

    ================================================== ==

    My 'solution' (of which I am most unsure!!!) is as follows:

    Idea generated by x + y, x - y is the ideal

     h_1 ( x + y) + h_2 (x - y) where  h_1, h_2 \in \mathbb{Q}[x,y]

    Ideal generated by x, y is the ideal

     h_3 x + h_4 y where  h_3, h_4 \in \mathbb{Q}[x,y]

    So

     h_1(x + y) + h_2 (x - y)  = h_1x + h_1y + h_2x - h_2y

     = (h_1 + h_2)x + (h_1 - h_2)y

     = h_3x + h_4y

     <x,y>

    Can someone please either correct this reasoning or confirm that is is correct/adequate.

    Peter
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  2. #2
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    Re: Ideals in Q[x,y]

    There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use \mathbb{Z}[x,y] instead?

    The most common way of proving two sets are equal in mathematics is to show they contain each other. Clearly, <x+y, x-y> is contained in the ideal generated by x and y, as both x+y and x-y are in <x,y> and <x+y,x-y> is the MINIMAL ideal containing x+y and x-y.

    So the "hard" part is showing that <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y>. And:

    x = (1/2)(x+y) + (1/2)(x-y) (note that we need to use rational numbers, integers wouldn't do!)

    y = (1/2)(x+y) - (1/2)(x-y), both of which are clearly elements of <x+y, x-y>.
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Ideals in Q[x,y]

    Quote Originally Posted by Deveno View Post
    There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use \mathbb{Z}[x,y] instead?

    The most common way of proving two sets are equal in mathematics is to show they contain each other. Clearly, <x+y, x-y> is contained in the ideal generated by x and y, as both x+y and x-y are in <x,y> and <x+y,x-y> is the MINIMAL ideal containing x+y and x-y.

    So the "hard" part is showing that <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y>. And:

    x = (1/2)(x+y) + (1/2)(x-y) (note that we need to use rational numbers, integers wouldn't do!)

    y = (1/2)(x+y) - (1/2)(x-y), both of which are clearly elements of <x+y, x-y>.

    Thanks Deveno, most helpful

    Having to think a bit about your statement "... <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y> ... but I will work on this ...

    Now, to address your question: " ... There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use \mathbb{Z}[x,y] instead?"

    ... In my 'proof' I really only showed that  <x + y, x - y> \ \  \subseteq  \ \ <x, y> since I showed the following

     f \in <x + y, x - y> \ \Longrightarrow \  f = h_1 (x + y) + h_2 (x - y) where  h_1, h_2 \in \mathbb{Q}

     \Longrightarrow \  f = h_3x + h_4y  \  \Longrightarrow \ f \in <x. y>

    so  <x + y, x - y> \ \ \subseteq \  \ <x, y>

    It appears that the above reasoning would work for  \mathbb{Z} [x,y] ... is that correct?'

    Following your proof for the second part (showing that <x,y> is contained in <x + y, x - y> ) this would not work for  \mathbb{Z} [x,y] since, as you point out, we need to use rational numbers.

    Thanks again for the help!

    Peter
    Last edited by Bernhard; July 31st 2013 at 07:29 AM.
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  4. #4
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    Re: Ideals in Q[x,y]

    Try proving a more general statement: if an ideal I of a ring R contains the set S, then I contains <S>, the ideal generated by S.
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