Do Irreducibles Induce Algebraic Extensions?

  • Thread starter slamminsammya
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In summary, the conversation discusses the relationship between a field K and the field extension K[x]/<f>, where f is an irreducible polynomial in K[x]. It is noted that in general, if M is a maximal ideal in a commutative ring R, then R/M is a field, and if R contains a subfield k, then R/M is an extension of k. In the case of K[x]/<f>, the ring K[x] contains the subfield K, and since K[x] is a Euclidean ring and a p.i.d., an irreducible polynomial f generates a maximal ideal and K[x]/<f> is a field extension of K. It is also mentioned that the degree of the extension
  • #1
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]?
 
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  • #2
i think so. in general, if M is a maximal ideal in a commutative ring R, then R/M is a field. If R contains a subfield k, then R/M is an extension of k. So in your case the ring K[X] contains the subfield K, and since K[X] is a Euclidean ring, hence also a p.i.d., an irreducible polynomial f generates a maximal ideal, so K[X]/(f) is a field extension of K.

Moreover the degree (vector dimension) of the extension equals the degree of f, hence is finite, and every finite extension is definitely algebraic. so YES!

I had to think through all the details since I am old and losing my memory. hope this helps.
 
  • #3
Isn't the answer obvious from the definition of "algebraic extension"? The interesting part is that it's a field and not just a ring.

(Hrm. I suppose there are equivalent definitions, and some would be less obvious than others. Which are you using?)
 

1. What are irreducibles in algebraic extensions?

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.

2. How do irreducibles induce algebraic extensions?

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.

3. What role do irreducibles play in Galois theory?

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.

4. How are irreducibles related to algebraic closure?

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.

5. Can all algebraic extensions be induced by irreducible polynomials?

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.

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