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I have the following definition of what it means for a polynomial to be separable:

Let $E$ be a field and $P \in E[x]$ be irreducible. Then we say $P$ is separable if it has no repeated root in any field extension of $E$. If $P\in E[x]$ is reducible, we say $P$ is separable if all of its irreducible factors are separable.

In the irreducible case, we have shown $P$ is separable if it has no repeated root in a splitting field extension.

Can anyone give me an example of a polynomial which is not separable? I don't see why this definition isn't trivial. For example, if I had a double root like $(x-2)^2$ then the polynomial is reducible so we consider $(x-2)$ and $(x-2)$ separately, each of which does not have a double root. So when could we ever have a polynomial which is not separable?

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up vote 13 down vote accepted

The typical example of a polynomial that is not separable: let $F=\mathbb{F}_p(x)$ be the field of all rational functions with coefficients in the field with $p$ elements. Consider the polynomial $y^p-x$ in $F[y]$. If $u$ is a root of $y^p-x$ (in some extension of $F$), we have $u^p-x=0$, so $u^p=x$. Because we are in charactersitic $p$, the binomial theorem becomes $$(a\pm b)^p = a^p \pm b^p,$$ so therefore we have $$(y-u)^p = y^p-u^p = y^p-x,$$ so $u$ is the only root of $y^p-x$ (by unique factorization), which is a perfect $p$ power in its splitting field.

Note also that $y^p-x$ is irreducible over $F$ (so that $u$ actually "lives" in some extension strictly larger than $F$). To prove this, consider e.g. by Eisenstein's criterion applied to it as an element of $(\mathbb{F}_p[x])[y]$: $\mathbb{F}_p[x]$ is a UFD where $x$ is prime; every coefficient of $y^p-x$ except the leading one is a multiple of $x$, and the constant term is not divisible by $x^2$; so by Eisenstein's Criterion $y^p-x$ is irreducible in $(\mathbb{F}_p[x])[y]$. Now Gauss's Lemma lets us conclude that the polynomial is also irreducible over $F$.

Over a field of characteristic $0$, all irreducible polynomials are separable. This follows because a polynomial has a multiple root if and only if the gcd with its formal derivative is not equal to $1$. But if $p(x)$ is irreducible, then $p'(x)$ is a polynomial of strictly smaller degree than $p(x)$, and therefore no nonconstant factor of $p(x)$ can be a factor of $p'(x)$. Thus, $\gcd(p(x),p'(x))=1$, so $p(x)$ is separable.

So inseparability is a purely positive characteristic phenomenon. In fact, one can show that an irreducible polynomial over a field of characteristic $p$ is not separable if and only if it can be written as a polynomial in $x^p$ (that is, every exponent that occurs is a multiple of $p$).

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I don't quite follow. How is $y^p-x$ irreducible if you just factored it as $(y-u)^p$? Do you have a more concrete example? – nullUser May 19 '12 at 21:01
@nullUser: The polynomial is irreducible over $F$. The factorization $(y-u)^p$ involves $u$ which is not in $F$. – Ted May 19 '12 at 21:06
@Ted: Oops; didn't read that comment carefully enough. Thanks. – Arturo Magidin May 19 '12 at 21:09
@nullUser: Sorry, I wasn't very clear. If $u$ is a root of the polynomial, possibly in some extension of $F$, then we see that $y^p-x$ is equal to $(y-u)^p$. Then we can prove (using Eisenstein's Criterion and Gauss's Lemma) that the polynomial is in fact irreducible over $F$, so that its roots do indeed "live" in some extension of $F$. So the polynomial is irreducible when considered as a polynomial in $F[y]$, but it has multiple roots when considered as a polynomial in $K[y]$, where $K$ is any field that contains roots for it. – Arturo Magidin May 19 '12 at 21:11

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