Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F\subset K$ be an algebraic field extension. Is the set of all elements of $K$ that are purely inseparable over $F$ necessarily a subfield of $K$?

share|improve this question

2 Answers 2

up vote 8 down vote accepted

Yes. An element $a\in K$ is purely inseparable iff $a^{p^n}\in F$ for some $n\geq 0$ (see here). Let $E$ be the subset of purely inseparable elements of $K$. For any $a,b\in E$, we have $a^{p^n}\in F$ and $b^{p^m}\in F$ for some $n,m\geq0$, so $$(-a)^{p^n}\in F\hskip 0.5in (a^{-1})^{p^n}=1/(a^{p^n})\in F$$ $$(a+b)^{p^{n+m}}=a^{p^{n+m}}+b^{p^{n+m}}=(a^{p^n})^{p^m}+(b^{p^m})^{p^n}\in F$$ and $$(ab)^{p^{n+m}}=a^{p^{n+m}}b^{p^{n+m}}=(a^{p^n})^{p^m}(b^{p^m})^{p^n}\in F$$ Thus, $E$ is a subfield.

share|improve this answer

Yes. This is clearer if you use the following definition of pure inseparability: an element $\alpha$ of an algebraic extension $F/K$ is purely inseparable if $|\text{Hom}_K(K(\alpha), \bar{K})| = 1$. Now it is obvious that if $\alpha, \beta$ are purely inseparable then any $K$-homomorphism $K(\alpha, \beta) \to \bar{K}$ is determined by what it does to $\alpha$ and $\beta$, so is unique.

share|improve this answer
    
The $\hom$ is the set of $K$-linear field homomorphisms, I guess :) –  Mariano Suárez-Alvarez Mar 13 '11 at 21:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.