# Extension fields / splitting fields proof...

• Dec 19th 2007, 06:50 AM
ElGamal
Extension fields / splitting fields proof...
Hello, put this in the homework help but think its probably best off here, Im trying to prove the following by induction....

Given a polynomial, f(X) \in F[X], of degree n, there exists
an extension field K (sub set of) F such that f(X) has n roots
in K.

This is what I did first time round...

First assume that F is irreducible and argue by induction on n.
Suppose the result holds for irreducible polynomials of degree at
most n-1. Set E= F[x]/f(x).
Now, in E, f(x) has at least one root, so we can write f(x)
= (x - \alpha_{1})(x - \alpha_{2})...(x-\alpha_{r})g(x) with
\alpha_{1}, \alpha_{2}, ..., \alpha_{r} \in E and g(x) \in
E[x] irreducible. Since deg g < n, by the inductive assumption,
the result also applies to g.

But ive been told this is wrong because there could be more than one irreducible factor of degree > 1, also I shouldnt be trying to take irreduciblity through the proof.

If someone knows how to prove this it would be much appreciated, its part of a huge project and needs to be done by tomorrow!

Cheers!

ElGamal.
• Dec 19th 2007, 07:29 AM
ThePerfectHacker
Quote:

Originally Posted by ElGamal
Hello, put this in the homework help but think its probably best off here, Im trying to prove the following by induction....

Given a polynomial, f(X) \in F[X], of degree n, there exists
an extension field K (sub set of) F such that f(X) has n roots
in K.

You need to use something call Kronecker's theorem. It is not hard, so we will prove it. Let F be a field and f(x) in F[x] be a non-constant polynomial. Then there exists an extension field K over F and $\displaystyle \alpha \in K$ such that $\displaystyle f(\alpha) = 0$. By unique factorization we can write f(x) as a product of irreducible polynomials. Thus it is necessary an sufficient that only of those factors p(x) has a zero $\displaystyle \alpha \in K$. Since $\displaystyle p(x)$ is a non-constant polynomial which is irreducible over F it means $\displaystyle F / \left< p(x) \right>$ is a field. Let us call this field $\displaystyle K$. We can identify that K contains F (up to isomorphism) by $\displaystyle \phi (a) = [a]_{p(x)}$ (where $\displaystyle [a]_{p(x)}$ is the equivalence class mod $\displaystyle p(x)$) for $\displaystyle a\in F$. Next we note that $\displaystyle [x]_{p(x)}$ is a zero of $\displaystyle p(x)= a_nx^n+...+a_10+a_0$ for $\displaystyle p([x]_{p(x)}) = [a_nx^n+...+a_1x+a_0]_{p(x)} = [0]_{p(x)}$. Thus $\displaystyle K$ is an extension field that we desired. Now to go into induction you have to note that if f(x) is an arbitrary non-constant polynomial then we factorize it $\displaystyle p_1(x)...p_k(x)$ were all $\displaystyle p_i(x)$ are irreducible. Now apply the argument on every irreducible factor $\displaystyle k$ times. Thus, there exists an extension field K such that $\displaystyle f(x)$ splits (meaning factors into linear factors).

You can make this result a little stronger: Given field F there is extension field K for a non-constant polynomial such such [K:F]<=n! where n = deg f(x) and f(x) splits over K.