# f split => f' split

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Mar 28, 2005, 9:00:12 PM3/28/05
to
Can anything interesting be said about a field K such that, for any
polynomial f over K which is split (all its root are in K), f' is also
split? Does this property have a name?

If K is real-closed, it is easy to see that it satisfies the property.
I'm not sure I can come up with any other example, in fact.

What about the smallest subfield of the algebraic closure of the
rationals which satisfies the property in question (it is clear that
it exists)? What does it "look like"?

--
(david....@ens.fr,

### G. A. Edgar

Mar 29, 2005, 9:30:05 AM3/29/05
to
In article <d2acrc$fi1$1...@news.ks.uiuc.edu>, David Madore
<david....@ens.fr> wrote:

> Can anything interesting be said about a field K such that, for any
> polynomial f over K which is split (all its root are in K), f' is also
> split? Does this property have a name?
>
> If K is real-closed, it is easy to see that it satisfies the property.
> I'm not sure I can come up with any other example, in fact.
>
> What about the smallest subfield of the algebraic closure of the
> rationals which satisfies the property in question (it is clear that
> it exists)? What does it "look like"?

An old Monthly problem is relevant.
Problem 5861, AMM 1972, p. 667.
Let F be an ordered field. [...] (b) If Rolle's Theorem holds in F,
does it follow that F is real-closed?
Here "Rolle's Theorem" means: If a polynomial f has zeros a<b, then f'
has a zero between a and b.

Solution to part (b), Feb. 1981, p. 150-152, by M. J. Pelling.
This solution constructs an ordered field where Rolle's Theorem holds,
but it is not real-closed.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

### steve fisk

Mar 29, 2005, 2:00:07 PM3/29/05
to
> Can anything interesting be said about a field K such that, for any
> polynomial f over K which is split (all its root are in K), f' is also
> split? Does this property have a name?
>
> If K is real-closed, it is easy to see that it satisfies the property.
> I'm not sure I can come up with any other example, in fact.
>
> What about the smallest subfield of the algebraic closure of the
> rationals which satisfies the property in question (it is clear that
> it exists)? What does it "look like"?
>

MR1903515 (2003b:12003)
Ballantine, C.(1-DTM); Roberts, J.(1-BOWD)
A simple proof of Rolle's theorem for finite fields.
Amer. Math. Monthly 109 (2002), no. 1, 72--74.
12E20 (11T06)

References: 0 Reference Citations: 1 Review Citations: 0
Using terminology drawn from a consequence of Rolle's theorem for the
real field, the authors say that an arbitrary field $F$ obeys Rolle's
property if the formal derivative of any polynomial in $F[x]$ that
splits completely over $F$ also splits completely over $F$. Kaplansky
asked for a classification of such fields, and T. Craven and\ G. Csordas
\ref[Illinois J. Math. 21 (1977), no. 4, 801--817; MR0568321 (58
\#27921)] and Craven \ref[Proc. Amer. Math. Soc. 125 (1997), no. 11,
3147--3153; MR1401731 (97m:12010)] carried out general investigations
that brought the particular conclusion that the finite fields that
obeyed Rolle's property were precisely those with 2 or 4 elements.
Claiming the approach of these papers to be technical, the authors now
provide an elementary proof of the finite field result.

\{Reviewer's remarks: It seems to the reviewer that an equally simple
proof is inherent in Craven's paper, especially Theorems 2.2 and 2.5.
For finite $F$ of odd characteristic $p$, consider $x^{p-1}(x+1)(x+a)$,
where $a$ is a non-square in $F$. For $p=2$, the crucial fact that, if
$F$ has Rolle's property, then the set $\{\gamma ^2 + \gamma \colon \gamma \in F \}$ is closed under multiplication (and so a subfield for
finite $F$), is already in Craven. Also, the paper's title is somewhat
misleading in that its main thrust is that even the weak version of
Rolle's theorem discussed is invalid except in two small fields.\}

### Alexander Borisov

Apr 2, 2005, 7:28:20 AM4/2/05
to

"David Madore" <david....@ens.fr> wrote in message
news:d2acrc$fi1$1...@news.ks.uiuc.edu...

> Can anything interesting be said about a field K such that, for any
> polynomial f over K which is split (all its root are in K), f' is also
> split? Does this property have a name?
>
...

>
> What about the smallest subfield of the algebraic closure of the
> rationals which satisfies the property in question (it is clear that
> it exists)? What does it "look like"?
>
> --
> David A. Madore
> (david....@ens.fr,
>
This subfield consists of the algebraic numbers with all real conjugates.
A simple proof of it is conatained in my paper:

MR1998173 (2004f:11136)
Borisov, Alexander(1-PAS)
On a question of Craven and a theorem of Belyi. (English. English summary)
Proc. Amer. Math. Soc. 131 (2003), no. 12, 3677--3679 (electronic).

Alexander Borisov
http://www.math.psu.edu/borisov/