splitting field of a finite set of polynomials

Lemma 1.

(Cauchy,Kronecker) Let $K$ be a field. For any irreducible polynomial $f$ in $K[X]$ there is an extension field of $K$ in which $f$ has a root.

Proof.

If $I$ is the ideal generated by $f$ in $K[X]$ , since $f$ is irreducible, $I$ is a maximal ideal of $K[X]$ , and consequently $K[X]/I$ is a field.
We can construct a canonical monomorphism $v$ from $K$ to $K[X]$ . By tracking back the field operation on $K[X]/I$ , $v$ can be extended to an isomorphism $w$ from an extension field $L$ of $K$ to $K[X]/I$ .
We show that $\alpha=w^{-1}(X+I)$ is a root of $f$ .
If we write $f=\sum_{i=1}^{n}f_{i}X^{i}$ then $f+I=0$ implies:

	$\displaystyle w(f(\alpha))$	$\displaystyle=w(\sum_{i=1}^{n}f_{i}\alpha^{i})$
		$\displaystyle=\sum_{i=1}^{n}w(f_{i})w(\alpha)^{i}$
		$\displaystyle=\sum_{i=1}^{n}v(f_{i})w(\alpha)^{i}$
		$\displaystyle=\sum_{i=1}^{n}(f_{i}+I)(X+I)^{i}$
		$\displaystyle=(\sum_{i=1}^{n}f_{i}X^{i})+I$
		$\displaystyle=f+I=0,$

which means that $f(\alpha)=0$ .∎

Theorem 1.

Let $K$ be a field and let $M$ be a finite set of nonconstant polynomials in $K[X]$ . Then there exists an extension field $L$ of $K$ such that every polynomial in $M$ splits in $L[X]$

Proof.

If $L$ is a field extension of $K$ then the nonconstant polynomials $f_{1},f_{2},...,f_{n}$ split in $L[X]$ iff the polynomial $\prod_{i=1}^{n}f_{i}$ splits in $L[X]$ . Now the proof easily follows from the above lemma. ∎

Title	splitting field of a finite set of polynomials
Canonical name	SplittingFieldOfAFiniteSetOfPolynomials
Date of creation	2013-03-22 16:53:09
Last modified on	2013-03-22 16:53:09
Owner	polarbear (3475)
Last modified by	polarbear (3475)
Numerical id	16
Author	polarbear (3475)
Entry type	Theorem
Classification	msc 12F05