Why isn't the perfect closure separable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:30:08Z http://mathoverflow.net/feeds/question/66178 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66178/why-isnt-the-perfect-closure-separable Why isn't the perfect closure separable? Bruno Stonek 2011-05-27T12:27:39Z 2011-08-09T15:33:32Z <p>This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.</p> <p>Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.</p> <p><a href="http://en.wikipedia.org/wiki/Separable_extension" rel="nofollow">Wikipedia</a>, following Isaacs, <em>Algebra, A Graduate Course</em> p.301, says:</p> <blockquote> <p>On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.</p> </blockquote> <p>The question is: <em>why?</em> And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But <em>why doesn't it?</em></p> <p>To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.</p> <p>ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered <a href="http://math.stackexchange.com/questions/36296/why-isnt-the-perfect-closure-separable" rel="nofollow">there</a> also, providing further intuition on this subject. </p> http://mathoverflow.net/questions/66178/why-isnt-the-perfect-closure-separable/66185#66185 Answer by unknown (google) for Why isn't the perfect closure separable? unknown (google) 2011-05-27T13:09:46Z 2011-05-28T17:44:28Z <p>In the special case that your extension $K/F$ is of the form $K=F(\text{root of }f)$ for some irreducible polynomial $f(x)$, then there is some nice intuition as follows. If $f$ is separable, then $K/F$ is separable, and nothing interesting is going on. If $f$ is inseparable, then that means $\gcd(f(x),f'(x))$ is not $1$. But $f(x)$ is irreducible, and the degree of $f'(x)$ is smaller than that of $f(x)$, so we must have $f'(x)\equiv 0$. This is only possible if $\operatorname{char}K=p$ and $f(x)=g(x^p)$ for some prime $p$ (just look termwise at the derivative and this should be obvious). Since $f$ is irreducible, so is $g$. Now, we continue this process to the polynomial $g$, until we have written $f(x)=h(x^{p^n})$ for some $n\geq 0$ and some separable polynomial $h$. Then the field extension is $F\subseteq F(\text{root of }h)\subseteq F((\text{root of }h)^{1/p^n})=K$, where the first part is separable, and the second is purely inseparable.</p> <p>With the above proof in mind, the intuition you are looking for might be the following. We know purely inseparable extensions come from taking $p$th roots in characteristic $p$. In the whole extension, you may have taken the $p$th root of something not in the ground field, and so you'd better take the separable closure before looking for the purely inseparable part of your field extension.</p> http://mathoverflow.net/questions/66178/why-isnt-the-perfect-closure-separable/66242#66242 Answer by Georges Elencwajg for Why isn't the perfect closure separable? Georges Elencwajg 2011-05-27T23:03:55Z 2011-05-28T16:53:28Z <p><strong>A counterexample</strong><br> Take as ground field $F=\mathbb F_2(u,v)$ and consider the polynomial $f(X)=X^6+uvX^2+u\in F[X]$. This polynomial is irreducible by Eisenstein. Let $F\subset K$ be the extension obtained by adjoining a root $a$ of $f(X)$ to F, so that $K=F[a]$, $[K:F]=6$ and $f(a)=0$ .The element $a^2\in K$ has minimal polynomial over $F$ the <em>separable</em> (because of degree $3$) polynomial $g(X)=X^3+uvX+u$ .<br> From this follows that the separable closure of $F$ inside $K$ is $F^{sep}=F[a^2]$ and that the extension $F^{sep}=F[a^2] \subset K=F[a]$ is purely inseparable, as expected. </p> <p>And now comes the surprise: there are no elements in $K$ purely inseparable over $F$ !(except the elements of $F$, of course) and so $K$ is <em>not</em> separable over the purely inseparable closure $F^{perf}=F$ of $F$. The proof that an element of $r\in K$ purely inseparable over $F$ belongs to $F$ is easy, once you realize that such an element must satisfy $r^2\in F$ . [Expand $r$ in the $F$-basis $a^i, 0\leq i\leq5$ and calculate $r^2$ (easy in characteristic $2$ !), remembering that $f(a)=0$.You will see that $r$ was already in $F$.]<br> (Edit: Kevin's comment and <a href="http://mathoverflow.net/questions/15987/if-k-k-is-a-finite-normal-extension-of-fields-is-there-always-an-intermediate-fi" rel="nofollow">link</a> have reminded that I had already given analogous examples in characteristic $p>2$, and completely forgotten about them! The two answers now cover all positive characteristics.)</p> <p><strong>A positive result</strong><br> If $F\subset K$ is <em>normal</em> , then indeed $K$ is separable over $F^{perf}$, the extensions $F^{sep}$ and $F^{perf}$ are linearly disjoint over $F$ and the canonical map $F^{sep} \otimes_F F^{perf} \to K$ is an isomorphism. (As an aside observe that this is one of the rare cases where the tensor product of two fields is a field)</p>