Separable Polynomials - Dummit and Foote - Proposition 37 .... ....

[Math Amateur]

I am reading David S. Dummit and Richard M. Foote : Abstract Algebra ...

I am trying to understand the proof of Proposition 37 in Section 13.5 Separable and Inseparable Extensions ...The Proposition 37 and its proof (note that the proof comes before the statement of the Proposition) read as follows:
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In the above text by D&F we read the following:

"If \(\displaystyle p(x)\) were inseparable then we have seen that \(\displaystyle p(x) = q(x^p)\) for some polynomial \(\displaystyle q(x) \in \mathbb{F} [x]\). ... ... "I cannot understand exactly why this is true ...Can someone please explain exactly why, as D&F assert ...\(\displaystyle p(x)\) inseparable \(\displaystyle \ \ \Longrightarrow \ \ p(x) = q(x^p)\) for some polynomial \(\displaystyle q(x) \in \mathbb{F} [x] \)I cannot find where D&F establish exactly this implication ... but maybe there is some idea in the proof of Corollary 34 which reads as follows:
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Corollary 36 may also be relevant ... so I am providing that as well ... as follows:
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Hope someone can help explain the assertion by D&F mentioned above ..

Peter

 

[jvicens]

Dear Peter,

Thank you for reaching out and asking for clarification on this topic. I am happy to help you understand the proof of Proposition 37 in Section 13.5 of "Abstract Algebra" by Dummit and Foote.

Firstly, let's define what it means for a polynomial to be inseparable. A polynomial p(x) over a field F is said to be inseparable if it cannot be factored into distinct linear factors in an extension field of F. In other words, if p(x) has a repeated root in its splitting field, then it is inseparable.

Now, let's look at the proof of Proposition 37. The proof starts by assuming that p(x) is inseparable. This means that p(x) cannot be factored into distinct linear factors in an extension field of F. Since p(x) is a polynomial, it can be written as a product of irreducible polynomials over F. Let's say that p(x) = f_1(x) f_2(x) ... f_n(x), where f_i(x) are irreducible polynomials over F.

Now, since p(x) is inseparable, it cannot be factored into distinct linear factors. This means that at least one of the irreducible factors, let's say f_1(x), must have a repeated root in its splitting field. This repeated root can be written as a polynomial q(x) = f_1(x^p), where p is the characteristic of the field F. This is because any element in the splitting field of f_1(x) can be written as a polynomial in its root, and since the root is repeated, the polynomial will have a degree less than p.

Therefore, we can write p(x) as p(x) = q(x) g(x), where g(x) = f_2(x) ... f_n(x). This shows that p(x) can be factored into two polynomials, q(x) and g(x), over F. And since q(x) is a polynomial in x^p, it is also a polynomial in x. This proves the statement p(x) = q(x^p) for some polynomial q(x) \in \mathbb{F} [x].

I hope this explanation helps you understand the proof of Proposition 37. If you have any further questions, please don't hesitate to ask.