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Thread: Splitting field

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    Senior Member vincisonfire's Avatar
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    Splitting field

    Show that, for any nonconstant $\displaystyle p(x) \in F(x) $, there is a field K containing F such that p(x) splits over K.
    I know that if q(x) is any nonconstant polynomial with coefficients from F , then there a field L containing F such that there is a root of q(x) in L.
    Now suppose p(x) is the product of irreducible polynomials $\displaystyle q_i(x) $ then we can take $\displaystyle L_i = F(x)/q_i(x) $ to be the field in which $\displaystyle q_i(x) $ has a root.
    My question is : would the direct sum of these fields be a field over which p(x) would split (supposing that in $\displaystyle L_i $, $\displaystyle q_i(x) $ splits )?
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    Quote Originally Posted by vincisonfire View Post
    Show that, for any nonconstant $\displaystyle p(x) \in F(x) $, there is a field K containing F such that p(x) splits over K.
    I know that if q(x) is any nonconstant polynomial with coefficients from F , then there a field L containing F such that there is a root of q(x) in L.
    Now suppose p(x) is the product of irreducible polynomials $\displaystyle q_i(x) $ then we can take $\displaystyle L_i = F(x)/q_i(x) $ to be the field in which $\displaystyle q_i(x) $ has a root.
    My question is : would the direct sum of these fields be a field over which p(x) would split (supposing that in $\displaystyle L_i $, $\displaystyle q_i(x) $ splits )?
    The main result is that if $\displaystyle f(x)$ is a non-constant polynomial in $\displaystyle F$ there exists an extension field $\displaystyle K$ such that there is $\displaystyle \alpha \in K$ so that $\displaystyle f(\alpha) = 0$. Given a non-constant polynomial $\displaystyle p(x)$ of degree $\displaystyle n$ you can construct $\displaystyle K_1$ so that $\displaystyle K_1$ is an extension field that $\displaystyle \alpha_1 \in K_1$ with $\displaystyle f(\alpha_1)=0$. Think of $\displaystyle p(x)$ as a polynomial in $\displaystyle K_1$ since it has $\displaystyle \alpha_1$ as a zero it means $\displaystyle p(x) = (x-\alpha_1)p_1(x)$ where $\displaystyle p_1(x)\in K_1[x]$. But that means there is a field $\displaystyle K_2$ so that $\displaystyle K_1\subseteq K_2$ and there is $\displaystyle \alpha_2\in K_2$ with $\displaystyle p_2(\alpha_2)=0$. And so this means $\displaystyle p(x) = (x-\alpha_1)(x-\alpha_2)p_2(x)$ if viewed as a polynomial in $\displaystyle K_2$. Continuing in this way we can construct, $\displaystyle F\subseteq K_1\subseteq K_2\subseteq ... \subseteq K_n$ so that the entire polynomial (which has degree $\displaystyle n$) can be written into linear factors over $\displaystyle K_n$.
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