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Math Help - Radical Ideals

  1. #1
    Senior Member slevvio's Avatar
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    Radical Ideals

    Hello all,

    I was wondering if anyone could help me answer the question: find ideals I,J of k[x_1,\ldots,x_n] such that \sqrt{I+J} \not= \sqrt{I} + \sqrt{J}, where k is an algebraically closed field (i.e. find a counterexample) Can anyone give me a hint about how to go about this? I tried messing around with \mathbb{C}[x] but I can't really work anything out.

    Any help would be appreciated, thank you
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  2. #2
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    Re: Radical Ideals

    Quote Originally Posted by slevvio View Post
    Hello all,

    I was wondering if anyone could help me answer the question: find ideals I,J of k[x_1,\ldots,x_n] such that \sqrt{I+J} \not= \sqrt{I} + \sqrt{J}, where k is an algebraically closed field (i.e. find a counterexample) Can anyone give me a hint about how to go about this? I tried messing around with \mathbb{C}[x] but I can't really work anything out.

    Any help would be appreciated, thank you
    in \mathbb{C}[x,y] choose I = (x) and J=(x^2+y^2). then \sqrt{I}=I because I is prime and \sqrt{J}=J because J is the intersection of two primes. so \sqrt{I}+\sqrt{J}=I+J=(x,y^2) but \sqrt{I+J}=\sqrt{(x,y^2)}=(x,y).
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  3. #3
    Senior Member slevvio's Avatar
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    Re: Radical Ideals

    Thank you,

    could you give me a bit more information about why \langle x^2 + y^2 \rangle is the intersection of 2 prime ideals?
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  4. #4
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    Re: Radical Ideals

    isn't \langle x^2 + y^2\rangle  = \langle x + iy\rangle \cap \langle x - iy\rangle ?
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  5. #5
    Senior Member slevvio's Avatar
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    Re: Radical Ideals

    I'm not sure. I'm very bad with ideals
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  6. #6
    Senior Member slevvio's Avatar
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    Re: Radical Ideals

    why is <x +iy> prime? I tried to show the quotient of C[x,y] by it is an intergral domain but to no avail
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  7. #7
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    Re: Radical Ideals

    x + iy is irreducible in \mathbb{C}[x,y] and we know that in a UFD, irreducible elements are prime and the ideal generated by a prime element is a prime ideal. another way is to define a homomorphism \varphi: \mathbb{C}[x,y] \to \mathbb{C}[y] by \varphi(f(x,y))=f(-iy,y) and then see that \ker \varphi = (x+iy).
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  8. #8
    Senior Member slevvio's Avatar
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    Re: Radical Ideals

    excellent thank you
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  9. #9
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    Re: Radical Ideals

    also, to see why (x^2+y^2)=(x+iy) \cap (x-iy), first we note that since x^2+y^2=(x+iy)(x-iy), it is clear that (x^2+y^2) \subseteq (x+iy) \cap (x-iy). for the converse, let z =(x+i)u = (x-iy)v for some u,v \in \mathbb{C}[x,y]. then x+iy divides (x-iy)v and thus x+iy divides v because \mathbb{C}[x,y] is a UFD and both x+iy and x-iy are irreducible. so v=(x+iv)w for some w \in \mathbb{C}[x,y] and hence z = (x^2+y^2)w \in (x^2+y^2).
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