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Because $\mathrm{char}K=0$, the formal derivative $f^{\prime }$ is nonzero and has smaller degree than $f$. Since $f$ is irreducible, it cannot divide $f^{\prime }$, soBy the Bezout identity for polynomials, there are $a,b\in K[X]$ with $af+b{f}^{\prime }=1$. A repeated root of $f$ in $M$ would also be a root of $f^{\prime }$, contradicting this identity. Thus the result that an irreducible polynomial in characteristic zero is separable gives

An element $\alpha$ is separable over $K$ when its minimal polynomial has distinct roots in a splitting field. A polynomial has a repeated root exactly when it has a common root with its formal derivative. Hence
$\gcd (f,f^{\prime })=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the minimal polynomial of $\beta$. Repeatedly factor through the Frobenius untilwhere $g^{\prime }\ne 0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha ={\beta }^{p^h}$ has separable minimal polynomial $g$. PutThen $M/K$ is separable andEvery $\gamma \in L=K(\beta )$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity givesso $L/M$ is purely inseparable.

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts ${\beta }^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.

In characteristic two, the formal derivative in positive characteristic ofisA polynomial is separable exactly when it and its derivative have no common root. If $r$ is a root of $f^{\prime }$, thenThus $f$ and $f^{\prime }$ have a common root exactly when $b=0$. Therefore $f$ is a separable polynomial if and only if $b\ne 0$.

For a ring homomorphism $A\to B$, the Module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.