Are irreducible polynomials square free?

Question

I have a half formed intuition that polynomials $f$ irreducible over a field $K$ have no repeated roots in its splitting field.

Is this true? What would be a proof of it?

Answer 1

I'll sum up the comments here : a field that has this property is called a perfect field. One says that irreducible polynomials over this field are separable, i.e. they split in linear factors (which is the same as saying they're squarefree). Fields of characteristic $0$ are perfect fields, and so are finite fields (for different reasons though). But there are other fields that are perfect ! (for starters, any algebraically closed field, since in them irreducible is equivalent to "of degree $1$").

To see this, note that if $K$ is of characteristic $0$, and $P = Q^2R$ then $P' = Q(QR' + 2Q'R)$, and so $Q \mid P\land P'$. $P$ being irreducible, $Q\mid P \implies Q \in K $ or $Q=P$. If $Q=P$, then $deg Q > deg P'$, which means $P'=0$ (as $Q\mid P'$), which, in characteristic $0$, means that $P\in K$: contradiction with the irreducibility.

In a finite field of characteristic $p$, $x\mapsto x^p$ is injective (it is an additive morphism and $x^p=0 \implies x=0$), so it is surjective (an injective map $f:A\to A$ is surjective whenever $A$ is finite). So if $P'= 0$, then $P = \displaystyle\sum_{k=0}^d a_k X^{pk}$ for some $a_k$'s and $d$, and if you find $b_k$ such that $b_k^p = a_k$ (which you can do, because the aforementioned map is surjective), then $P= \displaystyle\sum_{k=0}^d b_k^p X^{pk} = (\displaystyle\sum_{k=0}^d b_k X^{k})^p$ (in this last equality I used the equality called "the freshman's dream": in a field of characteristic $p$, $(a+b)^p = a^p + b^p$); but this means $P= Q^p$ for some $Q$, contradicting irreducibility. Therefore $P'\neq 0$, and so the same argument as above applies : $K$ is perfect.

Now notice that I didn't actually use the fact that $K$ was finite, rather I used the fact that $x\mapsto x^p$ was surjective; which is the case if $K$ is finite, but also happens on other occasions (as I said, it happens for instance in algebraically closed fields). Therefore, fields of characteristic $p$ in which this is true are also perfect. The converse is also true !

Indeed let $K$ be a field of characteristic $p$, and assume $y$ is not the $p$th power of any element of $K$. Consider $P= X^p - y$. Pick an extension $L/K$ in which $P$ has a root $x$. Then $P=X^p - x^p = (X-x)^p$ (freshman's dream again). Therefore if $P=QR, Q,R\in K[X]\setminus K$, $Q=(X-x)^s$ for some $0<s<p$.We then look at the $X^{s-1}$ coefficient of $Q$, which is $-sx$, which thus belongs to $K$. $0<s<p$ means $s\neq 0$ in $K$, therefore it is invertible : $x\in K$, which contradicts the hypothesis about $y$: from this we conclude that $P$ is irreducible; but also $P=(X-x)^p$ in $L$: $K$ is not perfect.

In conclusion, a field $K$ is perfect if and only if it is of characteristic $0$, or it is of characteristic $p$ and the map $x\mapsto x^p $ is surjective. (we often write this last condition more succintly $K^p = K$)