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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:
View attachment 6603
View attachment 6604
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
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:
View attachment 6603
View attachment 6604
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