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Math Help - Splitting fields

  1. #1
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    Splitting fields

    I'm trying to prove that splitting fields are unique up to isomorphism.

    Say \Sigma is a splitting field for a polynomial f over a field K. Suppose also that \Sigma' \supset K' is an isomorphic extension, ie that we have i:K\longrightarrow K' and j:\Sigma \longrightarrow \Sigma' isomorphisms with j|_{K} = i.

    How can I show that \Sigma' is a splitting field of i(f) over K'?

    I feel this should be intuitively obvious and that it should fall easily out of definitions and very basic facts, yet I can't seem to be able to prove it nicely.

    Thanks.
    Last edited by Aradesh; November 19th 2009 at 02:30 AM.
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  2. #2
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    Quote Originally Posted by Aradesh View Post
    I'm trying to prove that splitting fields are unique up to isomorphism.

    Say \Sigma is a splitting field for (a polynomial?) f over a field K. Suppose also that \Sigma' \supset K(K'?)is an isomorphic extension, ie that we have (an isomorphism?) i:K\longrightarrow K' and j:\Sigma \longrightarrow \Sigma' isomorphisms with j|_{K} = i.

    How can I show that \Sigma' is a splitting field of i(f) over K'?

    I feel this should be intuitively obvious and that it should fall easily out of definitions and very basic facts, yet I can't seem to be able to prove it nicely.

    Thanks.
    This is the sketch of the proof.

    Let f \in K[x] and use induction on n = [\Sigma: K].

    For n=1,\Sigma = K and f splits over K, which imples that i(f) splits over K'. Thus \Sigma'=K'.

    If n > 1, assume by induction that f \in K[x] of degree less than n holds the given isomorphism extension property. Then f must have an irreducible factor g of degree greater than 1. Let \alpha be a root of g in \Sigma such that f=gf_1, where f_1 has degree n-1. Verify that

    K(\alpha) \cong \frac{K[x]}{(g)} \cong \frac{K'[x]}{(g')} \cong K'(\alpha'), where i sends \alpha to \alpha' and g to g' with coefficients.

    Since \Sigma is a splitting field of f over K(\alpha) and \Sigma' is a splitting field of i(f) over K'(\alpha'), the induction hypothesis implies that j extends to an isomorphism \Sigma \cong \Sigma'.
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post
    This is the sketch of the proof.

    Let f \in K[x] and use induction on n = [\Sigma: K].

    For n=1,\Sigma = K and f splits over K, which imples that i(f) splits over K'. Thus \Sigma'=K'.

    If n > 1, assume by induction that f \in K[x] of degree less than n holds the given isomorphism extension property. Then f must have an irreducible factor g of degree greater than 1. Let \alpha be a root of g in \Sigma such that f=gf_1, where f_1 has degree n-1. Verify that

    K(\alpha) \cong \frac{K[x]}{(g)} \cong \frac{K'[x]}{(g')} \cong K'(\alpha'), where i sends \alpha to \alpha' and g to g' with coefficients.

    Since \Sigma is a splitting field of f over K(\alpha) and \Sigma' is a splitting field of i(f) over K'(\alpha'), the induction hypothesis implies that j extends to an isomorphism \Sigma \cong \Sigma'.
    Did you perform induction by considering [K(\alpha):\Sigma] < [K:\Sigma]=n and thus the field extensions K(\alpha)\subset\Sigma and K'(\alpha')\subset\Sigma' are isomorphic by induction?

    By the way I wasn't specifically asking for a proof of uniqueness of splitting fields, so I guess that caused some confusion. But this is still helpful

    What I meant though was this:
    I'm already assuming \Sigma \cong \Sigma', and I want to show that \Sigma' must be a splitting field for i(f). (I'm sure its a very trivial result)
    Last edited by Aradesh; November 19th 2009 at 03:46 AM.
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  4. #4
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    Actually I have figured out how to show that. Does this sound right?

    Let \alpha\in\Sigma be a root of f. Then (t-\alpha)|f \Rightarrow j(t-\alpha) | j(f)=i(f). Thus j(\alpha) is a root of i(f). So i(f) splits over \Sigma' and we can show that this is indeed the splitting field as if L' \subseteq \Sigma' is the splitting field of i(f) over K', then j^{-1}(\L')\subseteq \Sigma is a splitting field of f over K and therefore j^{-1}(L')=\Sigma=j^{-1}(\Sigma'), so L' must be all of \Sigma'.
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  5. #5
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    Actually I don't think what I did there is quite right as I didn't specifically show that every root of i(f) is contained in \Sigma'. How can i get around this? :/

    eg if f has distinct roots \{\alpha_1,...,\alpha_n\} then \{j(\alpha_1),\cdots,j(\alpha_n)\} are distinct and hence are all n roots of i(f), but if they're not distinct this argument breaks down.
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  6. #6
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    Quote Originally Posted by Aradesh View Post
    What I meant though was this:
    I'm already assuming \Sigma \cong \Sigma', and I want to show that \Sigma' must be a splitting field for i(f). (I'm sure its a very trivial result)
    Well, I think what I had shown in my previous post is an isomorphism between "splitting" fields using induction, which implies that \Sigma' is a splitting field for i(f).

    Zeroes of f in \bar{K} (algebraic closure of K) generate its splitting field \Sigma and the zeroes of i(f) in \bar{K'} generate its splitting field \Sigma' and we see that \alpha' is a root of i(f) in \bar{K'} and \alpha is a root of {i}^{-1}(if) in \bar{K}. By hypothesis of field isomorphism, the number of zeroes of f in \Sigma and the number of zeroes of i(f) in \Sigma' should be the same (Think of a vector space isomorphism !).

    If you want an alternative way of induction, this link might help you to construct inductive steps (here).
    Last edited by aliceinwonderland; November 19th 2009 at 01:55 PM.
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