Rupture field of an irreducible polynomial over a finite field equals its splitting field

Asked Dec 05 '21 at 14:34

Active Dec 06 '21 at 10:35

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Let $k= \mathbb{F}_{q}$ a finite field and $P \in k[X]$ an irreducible polynomial. Show that its rupture field is also its splitting field.

My take :

Let $K$ be the splitting field of P. By the primitive element theorem, $K= k(\alpha)$ for some $\alpha \in K$. If I show that $\alpha$ is a root of $P$, then it is finished. But is it a root of $P$ ?

edited Dec 06 '21 at 10:35

Servaes

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asked Dec 05 '21 at 14:34

Kieran McShane

What is a rupture field? And no, $\alpha$ need not be a root of $P$. – Servaes Dec 05 '21 at 22:36
@Servaes I mean by rupture field a field extension of $k$ generated by a root of $P$. – Kieran McShane Dec 05 '21 at 23:05
Thanks for the clarification. I have never heard it called that before. – Servaes Dec 05 '21 at 23:05
The question is handled as a part of this old question. I would expect there to be a better match. And won't use my dupehammer to point at a question I answered myself. – Jyrki Lahtonen Dec 07 '21 at 09:51
Another related oldie. – Jyrki Lahtonen Dec 07 '21 at 09:53
I want to also mention this. – Jyrki Lahtonen Dec 07 '21 at 09:55
This also works, assuming that you know the basic properties of normal extensions. – Jyrki Lahtonen Dec 07 '21 at 09:57
The same thing happens, when we know that the Galois group is abelian. The haul is a bit disappointin :-( – Jyrki Lahtonen Dec 07 '21 at 10:01

Rupture field of an irreducible polynomial over a finite field equals its splitting field

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