- #1

Math Amateur

Gold Member

MHB

- 3,998

- 48

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:

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

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:

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