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splitting field of a finite set of polynomials
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(Theorem)
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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:
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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"splitting field of a finite set of polynomials" is owned by polarbear.
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Cross-references: proof, iff, field extension, polynomials, finite set, implies, isomorphism, operation, monomorphism, canonical, maximal ideal, irreducible, ideal generated by, root, extension field, irreducible polynomial, field
This is version 13 of splitting field of a finite set of polynomials, born on 2007-04-01, modified 2007-06-03.
Object id is 9140, canonical name is AnyIrreduciblePolynomialOverAFieldHasARoot.
Accessed 1122 times total.
Classification:
| AMS MSC: | 12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions) |
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Pending Errata and Addenda
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