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Math Help - algebraic geometry

  1. #1
    Senior Member slevvio's Avatar
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    algebraic geometry

    Hello,

    Let k be a field of characteristic 2. Then find the ideal of X = V(t_1^2 + t_2^2 + t_3^2) \subseteq \mathbb{A}_k^3.

    I'm not sure how to solve this. I mean, since k is of characteristic two, I can see that  (t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2, so that I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}. Is it just what we'd think it would be, i.e. I_X = \langle t_1 + t_2 + t_3 \rangle ? If so, why is it true? But how do I find this radical ideal? And is this the best way to work out what the ideal is?

    Thanks for any help
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  2. #2
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    Re: algebraic geometry

    There is an 'easy' way to work out the radical of an ideal in case this is principal (which in your case it is):

    Let J=\langle f \rangle where f=\prod_{i=1}^n f_i^{k_i} is its factorization into irreducibles, then \sqrt{J} = \langle \prod_{i=1}^n f_i \rangle

    To prove this let g\in \sqrt{J} then there exists m such that g^m \in J ie. g^m = pf for some polynomial p, by the uniqueness of a factorization in a polynomial ring (over a field at least) the f_i must appear in the factorization of g, and so g\in \langle \prod_{i=1}^n f_i \rangle. On the other hand if g\in \langle \prod_{i=1}^n f_i \rangle then g^{\prod_{i=1}^n k_i} \in J.
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  3. #3
    Senior Member slevvio's Avatar
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    Re: algebraic geometry

    Thanks. Is there some way to ensure that the polynomial I have is definitely irreducible?
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  4. #4
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    Re: algebraic geometry

    Quote Originally Posted by slevvio View Post
    Hello,

    Let k be a field of characteristic 2. Then find the ideal of X = V(t_1^2 + t_2^2 + t_3^2) \subseteq \mathbb{A}_k^3.

    I'm not sure how to solve this. I mean, since k is of characteristic two, I can see that  (t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2, so that I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}. Is it just what we'd think it would be, i.e. I_X = \langle t_1 + t_2 + t_3 \rangle ? If so, why is it true? But how do I find this radical ideal? And is this the best way to work out what the ideal is?

    Thanks for any help
    I(V(J))=\sqrt{J} is not always true if your base field is not algebraically closed. is k algebraically closed?
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  5. #5
    Senior Member slevvio's Avatar
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    Re: algebraic geometry

    Apologies, the underlying assumption we use is that k is algebraically closed - forgot to mention this
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  6. #6
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    Re: algebraic geometry

    well then, the degree of each t_i in f =t_1+t_2+t_3 is one and so f has no non-trival factorization, i.e. f is irreducible. thus (f) is prime and so \sqrt{(f)}=(f). also \sqrt{J^m} = \sqrt{J}, for any ideal J and any integer m \geq 1. thus I(V((f^2)))=\sqrt{(f^2)}=\sqrt{(f)^2}=\sqrt{(f)}=(  f).
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  7. #7
    Senior Member slevvio's Avatar
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    Re: algebraic geometry

    Let's say that we have a field k which is not necessarily algebraically closed. Is it still true that V(I) = V(J) \iff \sqrt{I} = \sqrt{J}? For example, if this is true then I think that \sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle over any field k, but I am not too sure.
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  8. #8
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    Re: algebraic geometry

    Quote Originally Posted by slevvio View Post
    Let's say that we have a field k which is not necessarily algebraically closed. Is it still true that V(I) = V(J) \iff \sqrt{I} = \sqrt{J}? For example, if this is true then I think that \sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle over any field k, but I am not too sure.
    the answer to your first question is no. for example consider I =(x^2+y^2) and J=(x^4+y^4) over \mathbb{A}^2(\mathbb{Q}).
    the answer to your second question is yes, although you should have mentioned if you were considering \mathbb{A}^2(k) or \mathbb{A}^3(k), etc.
    you always need to give two important piece of infiormation: your ground field and the dimension.
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  9. #9
    Senior Member slevvio's Avatar
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    Re: algebraic geometry

    Ok, thanks. I should have written that I was working in affine 2-space. Unfortunately I worked out that result by assuming the result in the first question, so how would I calculate it properly. Do you have any recommendations of books I can read on this subject, because I am having real difficulty trying to work out radical ideals
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  10. #10
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    Re: algebraic geometry

    well, because x^2-1 \in (x-1), we have (x-1,x^2+y^2-1)=(x-1,y^2). now clearly (x-1,y) \subseteq \sqrt{(x-1,y^2)} and since (x-1,y) is a maximal ideal of k[x,y], we must have (x-1,y)=\sqrt{(x-1,y^2)}.

    it is a good exercise to prove that the same result holds, i.e. \sqrt{(x-1,y^2)}=(x-1,y), in affine n-space for any n \geq 2.

    note that the ideal (x-a_1, \ldots , x_n-a_n), for any field k and any a_i \in k, is always a maximal ideal of k[x_1, \ldots , x_n] because the map \varphi: k[x_1, \ldots . x_n] \longrightarrow k defined by \varphi(f(x_1, \ldots, x_n))=f(a_1, \ldots , a_n) is an onto ring homomorphism and \ker \varphi = (x_1-a_1, \ldots , x_n - a_n). if k is algebraically closed, then the converse is also true, by Nullstellensatz, i.e. every maximal ideal of k[x_1,x_2, \ldots , x_n] is in the form (x-a_1, \ldots , x_n-a_n) for some a_i \in k.

    there are algorithms for finding the radical of an ideal in polynomial rings but they are not easy and i don't think you'll need them. just simplify your ideal first and use the properties of radicals and remember what i just said about maximal ideals.
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  11. #11
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    Re: algebraic geometry

    In this case, as in many, you do not need to actually compute the radical of \mathfrak{p}. We have

    V=V(x^2+y^2+z^2)=V((x+y+z)^2)=V(x+y+z).

    But x+y+z irreducible \Rightarrow \mathfrak{p} prime
    \Rightarrow\, I(V(\mathfrak{p}))=\mathfrak{p}.
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