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The standard examples of irreducible, inseparable polynomials that one encounters in an introductory course on field theory all seem to have only a single root in an algebraic closure. Are there elementary examples of inseparable, irreducible polynomials with multiple different roots (at least one of which is repeated)? Equivalently, can a field extension contain elements which are inseparable, but whose minimal polynomials have more than one distinct root?

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Great question. I've taken an hour to think about it and still got no idea of an example. –  Patrick Da Silva Apr 1 '12 at 1:52
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Sure, examples can easily be written down. See Appendix A of www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable1.pdf, where an example is given, and from understanding that you can create your own examples. –  KCd Apr 1 '12 at 2:04
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Let me also point out that beyond just looking for an example, you should learn that all the roots of these mysterious polynomials you seek will have equal multiplicity as roots, and this multiplicity must be a power of $p$. There is more structure here than just random examples. –  KCd Apr 1 '12 at 5:15
    
Clickable links to Keith Conrad's handout: pdf document --- ktml page. –  Pierre-Yves Gaillard Apr 1 '12 at 10:52

3 Answers 3

up vote 9 down vote accepted

Let $p \in \mathbb N$ be prime, $q \in \mathbb N$ coprime to $p$, and let $F = \mathbb F_p(t)$ the field of rational functions of $t$ with coefficients in $\mathbb F_p$. Consider $$ f(x) = x^{pq} - t. $$ EDIT : By Eisenstein's criterion, $x^{pq} - t$ is irreducible over $\mathbb F_p[t]$ (because $t$ is a prime in there). By Gauss' Lemma, it is also irreducible over the field of fractions, which is $\mathbb F_p(t)$. Thanks to Sam L. for this part of my argument.

Since the derivative of $f$ is zero in $\mathbb F_p(t)[x]$, the polynomial is inseparable. But the polynomial $x^q - 1$ is separable in $\mathbb F_p(t)[x]$, because its derivative is $qx^{q-1}$, which has no common roots with $x^q - 1$, so that the roots of $x^q - 1$ are distinct. Now letting $\sqrt[pq]t$ be a root of $x^{pq} - t$ and $w$ a $q^{th}$ root of unity. Then the distinct roots of $f$ are $w^i (\sqrt[pq]t)$, with $i$ ranging from $0$ to $q-1$, each with multiplicity $p$.

Hope that helps,

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I know that it is "clearly" irreducible, but I can't seem to find out why. Any ideas, anyone? I think it has to do with the way I constructed things, i.e. that $t$ and $x$ must be algebraically independent or something. –  Patrick Da Silva Apr 1 '12 at 2:20
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Consider the ring $R= \mathbb F_p[t]\subset \mathbb F_p(t)$. $t$ is prime in $R$, so it follows from Eisensteins criterion that $X^{pq} - t$ is irreducible in $R$, now by Gauss' lemma it is also irreducible in $Q(R) = \mathbb F_p(t)$. –  Sam Apr 1 '12 at 2:29
    
@Sam L. : I was looking for such an argument. =) Thanks! I have only used Eisenstein's criterion over $\mathbb Z$ since I did a course in Galois theory, but I perfectly understand your point. It feels twisted though! –  Patrick Da Silva Apr 1 '12 at 2:39
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There is no reason to take $q$ to be prime here. It can be any integer greater than 1 that is not divisible by $p$. –  KCd Apr 1 '12 at 5:13
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@PatrickDaSilva Thanks that makes sense :D –  user38268 Apr 1 '12 at 9:41

EDIT: Discussions in the comments have convinced me to add a bit of introduction to my example, as follows: any separable extension of an inseparable extension ought to provide an example, so here's a very simple separable extension of a very simple inseparable extension:

Let $F={\bf F}_3(t)$, let $E=F(t^{1/3})$, let $K=E(\sqrt2)$. Note $[K:F]=[K:E][E:F]=2\times3=6$. Show that $K=F(t^{1/3}+\sqrt2)$ by showing that that element is not of degree 2 or 3. Then show that that element is what you're looking for.

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It is not quite clear at first glance why this polynomial satisfies the desired properties... I don't like your example because it doesn't give intuition ; it's not "transparent", compared to my example or Pierre-Yves'. –  Patrick Da Silva Apr 1 '12 at 8:36
    
@Patrick: Although this example is less simple than yours, it is a good example because it is created by a method other than the Eisenstein criterion. The minimal polynomial of $t^{1/3}+2$ over $F$ is $x^6 + tx^3 + (t^2+1)$. –  KCd Apr 1 '12 at 13:00
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For example, take $F = {\mathbf F}_p(t)$. For any positive integer $n$, $f(x) = x^n-t$ is irreducible over $F$ by Eisenstein and "therefore" $f(x^{p^r}) = x^{p^rn}-t$ is irreducible over $F$ for all $r\geq 1$, but that is easy to see directly by Eisenstein. I mention this first to show it includes your example. But it goes much further. Take $f(x) = x^2+tx+1$. This is irreducible in $F[x] = {\mathbf F}_p(t)[x]$ for any $p$ -- treat $p=2$ and $p > 2$ separately, perhaps -- and therefore $f(x^{p^r}) = x^{2p^r} + tx^{p^r} + 1$ is irreducible over $F$, inseparable, and it has two distinct roots. –  KCd Apr 1 '12 at 17:08
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More generally, if $f(x) \in {\mathbf F}_p(t)[x]$ is irreducible separable of degree $n$ and at least one of its $x$-coefficients is not a rational function in $t^p$, then $f(x^{p^r})$ is irreducible over ${\mathbf F}_p(t)$ for all $r \geq 1$, so we obtain a huge number of examples with $n$ distinct roots having common multiplicity as high as you wish (namely $p^r$) starting from a fairly flexible type of base case $f(x)$ -- much more flexible than $x^n-t$. Don't get me wrong: the binomial examples are good since they're simple, but it's also nice to see the process is much more general. –  KCd Apr 1 '12 at 17:12
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@Patrick, I took the simplest example I could of a separable extension of an inseparable extension. What could be simpler than extending by $\sqrt2$? And what simpler inseparable extension is there than ${\bf F}_3(t^{1/3})$? Well, I guess there's ${\bf F}_2(t^{1/2})$, but then I'd have to worry about what the extension by $\sqrt2$ looks like. So I took the simplest separable extension of the (almost-) simplest inseparable extension. Then I took the simplest possible generator, $t^{1/3}+\sqrt2$, of that extension. The only guessing involved was guessing that such a simple example would work. –  Gerry Myerson Apr 2 '12 at 2:55

Consider $$ (X-a)^p(X-b)^p\in\mathbb F_p(a^p+b^p,a^pb^p)[X], $$ where $a,b,X$ are indeterminates.

EDIT. A generalization: Let $q$ be a power of a prime, let $a_1,\dots,a_n$ be indeterminates, put $$ f:=(X-a_1)^{q^k}\cdots(X-a_n)^{q^k}\in\mathbb F_q[a_1,\dots,a_n,X], $$ write $K$ for the extension of $\mathbb F_q$ generated by the coefficients of $f$.

Then $f$ is irreducible in $K[X]$, and any example will be a specialization of this one.

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