Separable Polynomials - Remarks by Dummit and Foote .... ....

[Math Amateur]

Dummit and Foote in Section 13.5 on separable extensions make some remarks about separable polynomials that I do not quite follow. The remarks follow Corollary 34 and its proof ...

Corollary 34, its proof and the remarks read as follows:
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In the above text by D&F, in the remarks after the proof we read:

" ... in characteristic \(\displaystyle p\) the derivative of any power \(\displaystyle x^{pm}\) of \(\displaystyle x^p\) is identically \(\displaystyle 0\):

\(\displaystyle D_x( x^{pm} ) = pm x^{pm - 1 } = 0 \)

so it is possible for the degree of the derivative to decrease by more than \(\displaystyle 1\).

If the derivative \(\displaystyle D_x p(x)\) of the irreducible polynomial \(\displaystyle p(x)\) is non-zero, however, , then just as before we conclude that \(\displaystyle p(x)\) must be separable. ... ... "My questions are as follows:Question 1

I am assuming that when the degree of the derivative decreases by more than \(\displaystyle 1, p(x)\) is still relatively prime to \(\displaystyle D_x p(x)\) and so \(\displaystyle p(x)\) is separable ... is that the correct reasoning here ...Question 2

In stating that "if the derivative \(\displaystyle D_x p(x)\) of the irreducible polynomial \(\displaystyle p(x)\) is non-zero, however, , then just as before we conclude that \(\displaystyle p(x)\) must be separable". D&F are implying that if \(\displaystyle D_x p(x) = 0\) then \(\displaystyle p(x)\) is not separable (or not necessarily separable) ... is this the case ... if it is the case, why/how is this true ...
Hope that someone can help ...

Peter

 

[jvicens]

Dear Peter,

Thank you for bringing up these questions. it is important to clarify any doubts and uncertainties in order to fully understand a concept.

To answer your first question, yes, your reasoning is correct. When the degree of the derivative decreases by more than 1, p(x) is still relatively prime to D_x p(x), and therefore p(x) is separable. This is because if p(x) and D_x p(x) have a common factor, then p(x) would not be irreducible, contradicting the given information.

For your second question, you are correct in understanding that D&F are implying that if D_x p(x) = 0, then p(x) is not necessarily separable. This is because, as mentioned in the remarks, in characteristic p, the derivative of any power of x^p is identically 0. This means that if p(x) is a power of x^p, then its derivative would be 0 and it would not be separable. However, if p(x) is not a power of x^p, then its derivative would not be 0 and it would still be separable.

I hope this helps clarify the concept for you. If you have any further questions, please do not hesitate to ask. As scientists, it is important to fully understand and question concepts in order to advance our knowledge and understanding of the world. Keep up the good work!