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Separable extension - Encyclopedia of Mathematics
An irreducible polynomial $f(x)$ is inseparable if and only if its derivative $f'(x)$ is identically zero (this is
possible only for $k$ of characteristic $p>0$ and $f(x)=f_1(x^p)$).
The Stacks project 9.12 Separable extensions
Lemma 9.12.1. Let $F$ be a field. Let $P \in F[x]$ be an irreducible polynomial over $F$. Let $P' = \text{d}P/\text{d}x$ be the derivative of $P$ with respect to $x$. Then one of the following two cases happens
- $P$ and $P'$ are relatively prime, or
- $P'$ is the zero polynomial.
The second case can only happen if $F$ has characteristic $p > 0$. In this case $P(x) = Q(x^ q)$ where $q = p^ f$ is a power of $p$ and $Q \in F[x]$ is an irreducible polynomial such that $Q$ and $Q'$ are relatively prime.
Proof. Note that $P'$ has degree $< \deg (P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg (P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2).
Assume we are in case (2) and $P = a_ d x^ d + \ldots + a_0$. Then $P' = da_ d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_ d, \ldots , a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_ i = 0$ whenever $p$ does not divide $i$. In other words, $P(x) = P_1(x^ p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^ q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. $\square$ |
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楼主: 业余的业余
hbghlyj
发表于 2023-11-8 00:08
