# 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 |