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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.2012-04-01
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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.2012-04-01
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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.2012-04-01
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    Clickable links to Keith Conrad's handout: [pdf document](http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable1.pdf) --- [ktml page](http://www.math.uconn.edu/~kconrad/blurbs/).2012-04-01

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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.2012-04-01
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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](http://en.wikipedia.org/wiki/Eisenstein%27s_criterion) that $X^{pq} - t$ is irreducible in $R$, now by [Gauss' lemma](http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29) it is also irreducible in $Q(R) = \mathbb F_p(t)$.2012-04-01
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    @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!2012-04-01
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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 $p$.2012-04-01
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    @KCd : Yes, you are right. It just felt like a "smaller" example while I was still developing my argument. Of course it can be made more general. I edited my answer to show your comment.2012-04-01
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    @SamL. What does it mean for an indeterminate $t$ to be a prime element in $\Bbb{F}_p[t]$?2012-04-01
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    @Benjamin Lim : Given a ring $R$, a prime element of $R$ is such that if $p \, | \, ab$ (i.e. $\exists k \in R s.t. pk = ab$) , then $p \, | \, a$ or $p \, | \, b$. Now $t \in \mathbb F_p[t]$ is such an example, because if $t \, | \, p(t) q(t)$, then one of $p(t)$ or $q(t)$ has constant term zero, for if both of them don't, then $t$ doesn't divide... hence $t$ is prime.2012-04-01
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    @PatrickDaSilva Thanks that makes sense :D2012-04-01
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    what is $\mathbb F_p$?2015-03-08
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    @athos : The standard notation for the finite field with $p$ elements, where $p$ is a prime number.2015-03-10
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    @PatrickDaSilva thank you!2015-03-10
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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'.2012-04-01
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    @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 $x^6 + tx^3 + (t^2+1)$.2012-04-01
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    @Patrick: Another thing that is nice about Gerry's example is that if you think about what makes it work, you're led to the following result: if $F$ is any field of characteristic $p$ and $f(x)$ is irreducible in $F[x]$ and $f(x) \not\in F^p[x]$ (that is, the coefficients of $f(x)$ are not all $p$th powers in $F$ -- some coefficient in $f(x)$ is not a $p$th power in $F$) then $f(x^{p^r})$ is irreducible in $F[x]$ for every $r \geq 1$. This gives examples where the distinct roots have very high common $p$-power multiplicity without the crutch of having to use binomials $f(x) = x^n-a$.(contd...)2012-04-01
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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.2012-04-01
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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.2012-04-01
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    @KCd : I understand your point, but I only said I didn't like the example because it is not transparent ; of course it suggests more generality, but OP seemed to "wonder" if such polynomials existed, so examples such as Pierre-Yves' or mine were expected and are transparent. Your kind of argument is pushing it too much for someone who simply "wonders". But hey, I'm not judging ; the argument is cool, I just thought it wasn't appropriate as an answer.. but it's definitely interesting,.2012-04-01
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    @Patrick, "It is not quite clear at first glance why this polynomial satisfies the desired properties" - is that a bug or a feature? If it's not clear at first glance, then OP has to do a little work to dig it out, and that can't be a bad thing, can it? Actually, I was tempted to just say, "a separable extension of an inseparable extension is inseparable," and leave it at that, but I feared that might be leaving too much work for OP.2012-04-02
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    @Gerry Myerson : What I mean is that for instance, in my example, I've taken the $pq^{th}$ root of unity and added the (distinct) $q$^{th} roots of unity to construct my polynomial, so even if you don't go through the proof it makes sense that this is a good idea. Pierre-Yves' is an even better example in my point of view : he simply took a polynomial with linear factors and "created an extension out of it", which gives irreducibility for free. Perhaps your idea is good, but the way the answer is written, at first glance, it seems rather arbitrary. That's what I meant.2012-04-02
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    To sum it up, I feel that the idea behind your answer is not visible in it, whether you leave some blanks or not. I had to read the comments to understand it... "at first glance", I said "bleh, this feels like he just guessed stuff".2012-04-02
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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.2012-04-02
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    Perhaps that first sentence was all you needed to add to make things clear =) after giving a little thought about it, that's pretty much what you and I did ; my inseparable extension was given by the polynomial $x^p - \sqrt[p]t$ and its separable extension was by adding the $q^{th}$ roots of unity... you saw clear behind this trick and extracted the essential information out of it. Hm. Since I took a little more time to understand your idea and that I like it now, I'll give it a +1. =)2012-04-02
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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.