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.

This is written on page 4 of James E. Humphreys' Linear Algebraic Groups:

A derivation $\delta: E \rightarrow L$ ($E$ a field, $L$ an extension field of $E$), is a map which satisfies $\delta(x+y) = \delta(x)+\delta(y)$ and $\delta(xy)=x \delta(y)+ \delta(x)y$. If $F$ is a subfield of $E$, $\delta$ is called an $F$-derivation if in addition $\delta(x)=0$ for all $x \in F$ (so $\delta$ is $F$-linear).

Then the author said:

The space $\operatorname{Der}_F(E,L)$ of all $F$-derivations $E \rightarrow L$ is a vector space over $L$, whose dimension equals the transcendence degree of $E$ over $F$ if $E/F$ is separably generated. $E/F$ is separable if and only if all derivations $F \rightarrow L$ extend to derivations $E \rightarrow L$ ($L$ an extension field of $E$).

I am confused by the last statement. Let $F = F_5 = \{0,1,2,3,4 \} $, the field of $5$ elements, and $E$ the splitting field of $x^5 -2$ over $F$. For any extension field $L$ of $E$, any derivation $\delta: F \rightarrow L$ must be the zero map (because of the two conditions). So the zero map from $E$ to $L$ is the extension of $\delta$ to $E$. But, $E/F$ is obviously inseparable.

Where am I wrong? Are there any references or hints as to the proof of the statement?

Sincere thanks.


I was wrong because $F$ is perfect and the extension is trivial.

Now, please allow me to ask for some references or hints as to the proof of:

$E/F$ is separable if and only if all derivations $F\rightarrow L$ extend to derivations $E\rightarrow L$, where $L$ is an extension field of $E$.

share|improve this question
2  
Isn't $E = F$? In $F$ we have $2^5 = 2$. –  Dylan Moreland Jan 11 '12 at 19:33
    
@Dylan Moreland: Yes, you are right. Thanks very much for the comment. I was wrong. In fact, only infinite fields of prime characteristic could be imperfect... –  ShinyaSakai Jan 11 '12 at 19:39
    
@ShinyaSakai: "some references or hints as to the proof." The proof of what? That $F$ is perfect? That the extension is trivial? That algebraic extensions of perfect fields are always separable? Or that $E/F$ is separable if and only if all derivations extend? Please edit the question to clarify. –  Arturo Magidin Jan 11 '12 at 20:02
    
@Arturo Magidin: Thanks for the comment. I am looking for the proof of the statement that $E/F$ is separable if and only if all derivations extend. –  ShinyaSakai Jan 11 '12 at 20:07
    
@ShinyaSakai: Is it really too hard to be explicit and precise? Does "the last statement" refer to "the extension is trivial"? To "$F$ is perfect and the extension is trivial"? (Both are valid candidates for "the last statement"); or to final assertion of the second quote box (which is not "the last statement" made before the final sentence of the post)? –  Arturo Magidin Jan 11 '12 at 20:08

1 Answer 1

up vote 3 down vote accepted

Your error is in thinking that $E/F$ is inseparable.

First, note that $x^5-2$ is not irreducible: by Fermat's Little Theorem, $a^5 = a$ for all $a\in F$, so $2$ is a root of $x^5-2$ in $F$. Thus, the splitting field is actually the splitting field of an irreducible of degree less than $p$, and so will be separable (e.g., by the derivative test).

Also, note that finite fields are perfect (a field $k$ is perfect if and only if $k$ is of characteristic $0$, or $k$ is of characteristic $p$ and $k^p=k$). If $k$ is a perfect field, then every algebraic extension of $k$ is perfect. (E.g., Corollary V.6.12 in Lang's Algebra).

All finite fields are perfect, since the Frobenius map $x\mapsto x^p$ is an automorphism ($(a+b)^p = a^p+b^p$ and $(ab)^p =a^pb^p$; and $a^p=0$ if and only if $a=0$). So every algebraic extension of a finite field is necessarily separable.


The book you quote has three explicit references to the statement in question, yet you are asking for "references or hints"...

Lang's Algebra (Revised 3rd Edition), Chapter VIII, Section 5, page 370, has:

Let $K(x)$ be an extension of $K$, and let $D$ be a derivation of $K$. Let $f(X)$ be the irreducible polynomial satisfied by $x$ over $K$, and let $D$ be a derivation on $K$.

If $x$ is separable algebraic over $K$, let $f(X)$ be the irreducible polynomial of $x$ over $K$. Then $f'(x)\neq 0$, and if we define $$u = -\frac{f^D(x)}{f'(x)},$$ where $f^D$ is the polynomial obtained from $f$ by applying $D$ to the coefficients of $f$, then for any $g(x),h(x)\in K[x]$, $h(x)\neq 0$, we define $$\begin{align*} D^*g(x) &= g^D(x) + g'(x)u\\ D^*(g/h) = \frac{hD^*g - gD^*h}{h^2} \end{align*}$$ gives a derivation on $K(x)$ which is an extension of $D$.

If $x$ is transcendental over $K$, then you can extend $D$ by defining it as above with $u$ an arbitrary element of $K(x)$.

And if $x$ is purely inseparable over $K$, so $x^p-a=0$ for some $a\in K$, then $D$ extends to $K(x)$ if and only if $Da=0$; so if $D$ is trivial on $K$, you can extend it by selecting $u$ arbitrarily again.

Then we have:

Proposition 5.2. A finitely geenrated extension $K(x_1,\ldots,x_n)$ over $K$ is separable algebraic if and only if every derivation $D$ of $K(x_1,\ldots,x_n)$ which is trivial on $K$ is trivial on $K(x_1,\ldots,x_n)$.

Proof. If $K(x_1,\ldots,x_n)$ is separable, then one can extend by applying the first case above repeatedly. Is the extension is not separable, we can decompose it into a tower of extensions between $K$ and $K(x)$ such that each step is either separable, transcendental, or purely inseparable. At least one step must be of one of the latter two types. The the uppermost such step, and use it to construct a derivation which is trivial on the base but not on the top to establish the converse.

This gives you how to extend the derivation if $E/F$ is separable.

For the converse, note that if $K(x_1,\ldots,x_n)$ is a finitely generated extension, and $f\in K[X_1,\ldots,X_n]$, let $\partial f/\partial x_i$ be $\partial f/\partial X_i$ evaluated at $(x_1,\ldots,x_n)$. If $f(X)\in K[X_1,\ldots,X_n]$ is a polynomial that vanishes on $(x_1,\ldots,x_n)$, then any extension $D^*$ of $D$ must satisfy $$0 = D^*f(x) = f^D(x) + \sum\frac{\partial f}{\partial x_i}D^*x_i;$$ and this condition is also sufficient for an extension to exist. If $E/F$ is not separable, then use this necessary condition to construct a derivation that cannot be extended.

share|improve this answer
    
Thank you very much for the detailed explanation! –  ShinyaSakai Jan 11 '12 at 19:43
    
Thank you!! I don't have those books at hand, so thanks very much for typing and organizing all the materials. It is really very kind of you. –  ShinyaSakai Jan 13 '12 at 18:03
    
@Shinya: You should really give that information; as I said, the book you quote gives references, but all you do is ask "for references", which, frankly, would suggest to the casual reader that you are either too careless to notice or too lazy to look them up. If the problem is that the references given are not accessible to you, then say so explicitly when you request references and/or "hints". Say: "The book refers me to Lang's Algebra, or to blah-blah-blah, but I don't have access to those books" etc. –  Arturo Magidin Jan 13 '12 at 18:08
    
Thanks for all the kind help. I will try my best to be a better user of this site and this language. –  ShinyaSakai Jan 13 '12 at 19:23

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.