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Thread: Splitting fields

  1. #1
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    Splitting fields

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

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

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

    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; Nov 19th 2009 at 01:30 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 $\displaystyle \Sigma$ is a splitting field for (a polynomial?)$\displaystyle f$ over a field $\displaystyle K$. Suppose also that $\displaystyle \Sigma' \supset K$(K'?)is an isomorphic extension, ie that we have (an isomorphism?)$\displaystyle i:K\longrightarrow K'$ and $\displaystyle j:\Sigma \longrightarrow \Sigma'$ isomorphisms with $\displaystyle j|_{K} = i$.

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

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

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

    If n > 1, assume by induction that $\displaystyle f \in K[x]$ 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 $\displaystyle \alpha$ be a root of g in $\displaystyle \Sigma$ such that $\displaystyle f=gf_1$, where $\displaystyle f_1$ has degree n-1. Verify that

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

    Since $\displaystyle \Sigma$ is a splitting field of f over $\displaystyle K(\alpha)$ and $\displaystyle \Sigma'$ is a splitting field of i(f) over $\displaystyle K'(\alpha')$, the induction hypothesis implies that j extends to an isomorphism $\displaystyle \Sigma \cong \Sigma'$.
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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 $\displaystyle f \in K[x]$ and use induction on $\displaystyle n = [\Sigma: K]$.

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

    If n > 1, assume by induction that $\displaystyle f \in K[x]$ 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 $\displaystyle \alpha$ be a root of g in $\displaystyle \Sigma$ such that $\displaystyle f=gf_1$, where $\displaystyle f_1$ has degree n-1. Verify that

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

    Since $\displaystyle \Sigma$ is a splitting field of f over $\displaystyle K(\alpha)$ and $\displaystyle \Sigma'$ is a splitting field of i(f) over $\displaystyle K'(\alpha')$, the induction hypothesis implies that j extends to an isomorphism $\displaystyle \Sigma \cong \Sigma'$.
    Did you perform induction by considering $\displaystyle [K(\alpha):\Sigma] < [K:\Sigma]=n$ and thus the field extensions $\displaystyle K(\alpha)\subset\Sigma$ and $\displaystyle K'(\alpha')\subset\Sigma'$ 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 $\displaystyle \Sigma \cong \Sigma'$, and I want to show that $\displaystyle \Sigma'$ must be a splitting field for $\displaystyle i(f)$. (I'm sure its a very trivial result)
    Last edited by Aradesh; Nov 19th 2009 at 02:46 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 $\displaystyle \alpha\in\Sigma$ be a root of $\displaystyle f$. Then $\displaystyle (t-\alpha)|f \Rightarrow j(t-\alpha) | j(f)=i(f)$. Thus $\displaystyle j(\alpha)$ is a root of $\displaystyle i(f)$. So $\displaystyle i(f)$ splits over $\displaystyle \Sigma'$ and we can show that this is indeed the splitting field as if $\displaystyle L' \subseteq \Sigma'$ is the splitting field of $\displaystyle i(f)$ over $\displaystyle K'$, then $\displaystyle j^{-1}(\L')\subseteq \Sigma$ is a splitting field of $\displaystyle f$ over $\displaystyle K$ and therefore $\displaystyle j^{-1}(L')=\Sigma=j^{-1}(\Sigma')$, so $\displaystyle L'$ must be all of $\displaystyle \Sigma'$.
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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 $\displaystyle i(f)$ is contained in $\displaystyle \Sigma'$. How can i get around this? :/

    eg if $\displaystyle f$ has distinct roots $\displaystyle \{\alpha_1,...,\alpha_n\}$ then $\displaystyle \{j(\alpha_1),\cdots,j(\alpha_n)\}$ are distinct and hence are all $\displaystyle n$ roots of $\displaystyle i(f)$, 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 $\displaystyle \Sigma \cong \Sigma'$, and I want to show that $\displaystyle \Sigma'$ must be a splitting field for $\displaystyle i(f)$. (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 $\displaystyle \Sigma'$ is a splitting field for $\displaystyle i(f)$.

    Zeroes of f in $\displaystyle \bar{K}$ (algebraic closure of K) generate its splitting field $\displaystyle \Sigma$ and the zeroes of i(f) in $\displaystyle \bar{K'}$ generate its splitting field $\displaystyle \Sigma'$ and we see that $\displaystyle \alpha'$ is a root of $\displaystyle i(f)$ in $\displaystyle \bar{K'}$ and $\displaystyle \alpha$ is a root of $\displaystyle {i}^{-1}(if)$ in $\displaystyle \bar{K}$. By hypothesis of field isomorphism, the number of zeroes of f in $\displaystyle \Sigma$ and the number of zeroes of i(f) in $\displaystyle \Sigma'$ 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; Nov 19th 2009 at 12:55 PM.
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