splitting field of a finite set of polynomials

Lemma 1.

(Cauchy,Kronecker) Let $K$ be a field. For any irreducible polynomial $f$ in $K\mathit{}\mathrm{[}X\mathrm{]}$ 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:

$w(f(\alpha ))$ $=w({\displaystyle \sum _{i=1}^n}{f}_i{\alpha }^i)$

$={\displaystyle \sum _{i=1}^n}w(f_i)w{(\alpha )}^i$

$={\displaystyle \sum _{i=1}^n}v(f_i)w{(\alpha )}^i$

$={\displaystyle \sum _{i=1}^n}(f_i+I){(X+I)}^i$

$=({\displaystyle \sum _{i=1}^n}{f}_i{X}^i)+I$

$=f+I=0,$

which means that $f(\alpha )=0$.∎

Theorem 1.

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

Proof.

If $L$ is a field extension of $K$ then the nonconstant polynomials $f_1,f_2,\mathrm{\dots },f_n$ 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