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slamminsammya
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Given [tex]K[/tex] a field, and [tex] f\in K[x][/tex] an irreducible (monic) polynomial. Does it follow that the field [tex] K[x]/\left<f\right>[/tex] is an algebraic extension of [tex]K[/tex]?
Irreducibles are polynomials that cannot be factored into smaller polynomials with coefficients in the same field. In algebraic extensions, these polynomials are used to generate new elements that are not in the original field.
When a polynomial is irreducible, it can be used to generate a field extension by adding a root of the polynomial to the original field. This new element, along with the original field, forms a larger field that contains all the roots of the polynomial.
In Galois theory, irreducible polynomials play a crucial role in determining the structure of field extensions and their automorphisms. They are used to identify subfields and subgroups, and to determine whether a field extension is solvable by radicals.
In order to construct an algebraically closed field, we must adjoin all the roots of every irreducible polynomial in the original field. This is because irreducibles cannot be factored, so they represent elements that are missing in the original field.
No, not all algebraic extensions can be induced by irreducible polynomials. Some extensions, such as transcendental extensions, do not involve irreducible polynomials. Additionally, not all fields have algebraic closures, so not all extensions can be induced by irreducibles.