1. ## 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 )?

2. Originally Posted by vincisonfire
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 $\displaystyle K_n$.