Splitting Fields - Example 3 - D&F Section 13.4, pages 537 - 538

In summary, Example 3 in Section 13.4 of Dummit and Foote's Field Theory discusses the degree of the extension K over Q(\sqrt[3]{2}). The fact that \sqrt{-3} is not in Q(\sqrt[3]{2}) implies that the degree of the extension is greater than 1. However, since \sqrt{-3} is a root of x^2 + 3, its minimal polynomial has degree at most 2. This leads to the conclusion that the degree of the extension is exactly 2, as \sqrt{-3} must be included in the extension to satisfy the minimal polynomial. This ensures that Q(\sqrt[3]{2}) is not the splitting field
  • #1
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I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Section 13.4 : Splitting Fields and Algebraic Closures ... ...

I need some help with an aspect of Example 3 of Section 13.4 ... ...

Example 3 reads as follows:
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In the above text by Dummit and Foote, we read the following:

" ... ... Since \(\displaystyle \sqrt{ -3 }\) satisfies the equation \(\displaystyle x^2 + 3 = 0\) the degree of this extension over \(\displaystyle \mathbb{Q} ( \sqrt [3] {2} )\) is at most \(\displaystyle 2\), hence must be \(\displaystyle 2\) since we observed above that \(\displaystyle \mathbb{Q} ( \sqrt [3] {2} ) \) is not the splitting field ... ... "I do not understand why the degree of the extension \(\displaystyle K\) over \(\displaystyle \mathbb{Q} ( \sqrt [3] {2} )\) must be exactly \(\displaystyle 2\) ... ... why does \(\displaystyle \mathbb{Q} ( \sqrt [3] {2} )\) not being the splitting field ensure this ... ... ?

Can someone please give a simple and complete explanation ...

Hope someone can help ...

Peter
 
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  • #2
It could have been worded differently, but here is the point. Since $\sqrt{-3}\notin Q(\sqrt[3]{2}) := L$, then $[K:Q(\sqrt[3]{2})] = [L(\sqrt{-3}):L] > 1$. On the other hand, since $\sqrt{-3}$ is a root of $x^2 + 3$, its minimal polynomial has degree $\le 2$ and thus $[L(\sqrt{-3}):L] \le 2$. This forces $[L(\sqrt{-3}): L] = 2$, i.e., $[K:Q(\sqrt[3]{2})] = 2$.
 
  • #3
Euge said:
It could have been worded differently, but here is the point. Since $\sqrt{-3}\notin Q(\sqrt[3]{2}) := L$, then $[K:Q(\sqrt[3]{2})] = [L(\sqrt{-3}):L] > 1$. On the other hand, since $\sqrt{-3}$ is a root of $x^2 + 3$, its minimal polynomial has degree $\le 2$ and thus $[L(\sqrt{-3}):L] \le 2$. This forces $[L(\sqrt{-3}): L] = 2$, i.e., $[K:Q(\sqrt[3]{2})] = 2$.
Thanks Euge ...

just now reflecting on what you have said ...

Peter
 

FAQ: Splitting Fields - Example 3 - D&F Section 13.4, pages 537 - 538

What is a splitting field?

A splitting field is a field extension of a given field that contains all the roots of a given polynomial. It is the smallest field in which the polynomial can be fully factored into linear factors.

How do you find the splitting field of a polynomial?

To find the splitting field of a polynomial, you must first factor the polynomial into irreducible factors. Then, you can construct a field extension by adjoining all the roots of the polynomial to the original field. This field extension will be the splitting field.

Why is finding splitting fields important?

Finding splitting fields is important in the study of field theory because it allows us to fully factor polynomials and understand their roots. It also helps in solving equations, proving theorems, and constructing other field extensions.

Can a polynomial have multiple splitting fields?

Yes, a polynomial can have multiple splitting fields. However, all splitting fields of a polynomial will be isomorphic, meaning they have the same algebraic structure.

How do you know when you have found the splitting field of a polynomial?

You know you have found the splitting field of a polynomial when the polynomial can be fully factored into linear factors in the field extension. Additionally, the splitting field will contain all the roots of the polynomial and cannot be further extended.

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