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

[Math Amateur]

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

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

Hope someone can help ...

Peter

 

[Euge]

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., $[K:Q(\sqrt[3]{2})] = 2$.

 

[Math Amateur]

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