# Thread: Extension fields / splitting fields proof...

1. ## 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.

2. 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 $\alpha \in K$ such that $f(\alpha) = 0$. 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 $\alpha \in K$. Since $p(x)$ is a non-constant polynomial which is irreducible over F it means $F / \left< p(x) \right>$ is a field. Let us call this field $K$. We can identify that K contains F (up to isomorphism) by $\phi (a) = [a]_{p(x)}$ (where $[a]_{p(x)}$ is the equivalence class mod $p(x)$) for $a\in F$. Next we note that $[x]_{p(x)}$ is a zero of $p(x)= a_nx^n+...+a_10+a_0$ for $p([x]_{p(x)}) = [a_nx^n+...+a_1x+a_0]_{p(x)} = [0]_{p(x)}$. Thus $K$ 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 $p_1(x)...p_k(x)$ were all $p_i(x)$ are irreducible. Now apply the argument on every irreducible factor $k$ times. Thus, there exists an extension field K such that $f(x)$ 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.