# Splitting field

• Feb 19th 2009, 01:22 PM
vincisonfire
Splitting field
Show that, for any nonconstant $p(x) \in F(x)$, 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 $q_i(x)$ then we can take $L_i = F(x)/q_i(x)$ to be the field in which $q_i(x)$ 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 $L_i$, $q_i(x)$ splits )?
• Feb 19th 2009, 02:25 PM
ThePerfectHacker
Quote:

Originally Posted by vincisonfire
Show that, for any nonconstant $p(x) \in F(x)$, 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 $q_i(x)$ then we can take $L_i = F(x)/q_i(x)$ to be the field in which $q_i(x)$ 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 $L_i$, $q_i(x)$ splits )?

The main result is that if $f(x)$ is a non-constant polynomial in $F$ there exists an extension field $K$ such that there is $\alpha \in K$ so that $f(\alpha) = 0$. Given a non-constant polynomial $p(x)$ of degree $n$ you can construct $K_1$ so that $K_1$ is an extension field that $\alpha_1 \in K_1$ with $f(\alpha_1)=0$. Think of $p(x)$ as a polynomial in $K_1$ since it has $\alpha_1$ as a zero it means $p(x) = (x-\alpha_1)p_1(x)$ where $p_1(x)\in K_1[x]$. But that means there is a field $K_2$ so that $K_1\subseteq K_2$ and there is $\alpha_2\in K_2$ with $p_2(\alpha_2)=0$. And so this means $p(x) = (x-\alpha_1)(x-\alpha_2)p_2(x)$ if viewed as a polynomial in $K_2$. Continuing in this way we can construct, $F\subseteq K_1\subseteq K_2\subseteq ... \subseteq K_n$ so that the entire polynomial (which has degree $n$) can be written into linear factors over $K_n$.