A polynomial $p(x)\in F[x]$ is called separable if it has no multiple roots in its splitting field. Equivalently, if
\[p(x)=c(x-\alpha_1)\cdots(x-\alpha_n)\]in a splitting field, then $p(x)$ is separable if the roots $\alpha_1,\dots,\alpha_n$ are all distinct.
Example 1. The polynomial $x^2-2\in \mathbb{Q}[x]$ is separable because its roots are $\sqrt{2}$ and $-\sqrt{2}$, which are distinct. However, the polynomial $(x^2-2)^n$ is not separable for $n\ge 2$, because its roots occur with multiplicity $n$.
There is a simple criterion for detecting multiple roots. First, we define the derivative of a polynomial. If
\[p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\]then its derivative is
\[Dp(x)=nc_nx^{n-1}+(n-1)c_{n-1}x^{n-2}+\cdots+c_1\]Theorem. A polynomial $p(x)\in F[x]$ is separable if and only if
\[\gcd(p(x),Dp(x))=1\]Proof
We show that $p(x)$ has a multiple root if and only if $p(x)$ and $Dp(x)$ have a common root.
Suppose $\alpha$ is a multiple root of $p(x)$. Then in a splitting field, we can write
\[p(x)=(x-\alpha)^m g(x)\]with $m\ge 2$. Taking derivatives gives
\[Dp(x)=m(x-\alpha)^{m-1}g(x)+(x-\alpha)^mDg(x)\]Thus $\alpha$ is also a root of $Dp(x)$.
Conversely, suppose $\alpha$ is a root of both $p(x)$ and $Dp(x)$. Since $\alpha$ is a root of $p(x)$, write
\[p(x)=(x-\alpha)h(x)\]Then
\[Dp(x)=h(x)+(x-\alpha)Dh(x)\]Substituting $x=\alpha$ gives
\[Dp(\alpha)=h(\alpha)\]But $Dp(\alpha)=0$, so $h(\alpha)=0$. Hence $x-\alpha$ divides $h(x)$, and therefore $(x-\alpha)^2$ divides $p(x)$. Thus $\alpha$ is a multiple root.
So $p(x)$ has a multiple root if and only if $p(x)$ and $Dp(x)$ have a common root. Therefore $p(x)$ is separable if and only if
\[\gcd(p(x),Dp(x))=1\]For irreducible polynomials, this criterion becomes especially simple.
Theorem. Let $p(x)\in F[x]$ be irreducible. Then $p(x)$ is separable if and only if
\[Dp(x)\neq 0\]Proof
By the previous theorem, $p(x)$ is separable if and only if
\[\gcd(p(x),Dp(x))=1\]Since $p(x)$ is irreducible, the only possibilities are
\[\gcd(p(x),Dp(x))=1\]or
\[\gcd(p(x),Dp(x))=p(x)\]The second case means $p(x)$ divides $Dp(x)$. But $\deg Dp<\deg p$, so this can only happen if
\[Dp(x)=0\]Therefore $p(x)$ is not separable if and only if $Dp(x)=0$. Equivalently, $p(x)$ is separable if and only if
\[Dp(x)\neq 0\]This immediately gives the following theorem:
Theorem. If $F$ has characteristic $0$, then every irreducible polynomial in $F[x]$ is separable. Therefore every algebraic extension of $F$ is separable.
Proof
Let $p(x)\in F[x]$ be irreducible. Since $\operatorname{char}(F)=0$, the derivative of any nonconstant polynomial is not identically zero. Thus
\[Dp(x)\neq 0\]By the previous theorem, $p(x)$ is separable.
Therefore every algebraic element over $F$ has a separable minimal polynomial. Hence every algebraic extension of $F$ is separable.
Thus, algebraic extensions of fields such as $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are automatically separable. Inseparability can only occur in positive characteristic, below provides an example.
Example 2. Let $F=\mathbb{F}_p(t)$, where $t$ is transcendental over $\mathbb{F}_p$. Consider
\[p(x)=x^p-t\in F[x]\]If $\alpha$ is a root, then
\[\alpha^p=t\]In characteristic $p$, we have
\[Dp(x)=px^{p-1}=0\]Also,
\[x^p-t=x^p-\alpha^p=(x-\alpha)^p\]So $p(x)$ has only one root, with multiplicity $p$. Therefore $\alpha$ is algebraic over $F$, but not separable over $F$.