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Homesplitting field of a finite set of polynomials

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# 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. ∎

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## Corrections

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