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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 $K$?

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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.

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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.

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  • $\begingroup$ The $\hom$ is the set of $K$-linear field homomorphisms, I guess :) $\endgroup$ – Mariano Suárez-Álvarez Mar 13 '11 at 21:14

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