# Separable field extensions *without* using embeddings or automorphisms

If $K\subseteq L$ is a field extension and $x\in L$ is algebraic, we say that $x$ is separable over $K$ iff its minimal polynomial $f$ over $K$ is separable (i.e., $f$ is relatively prime with its derivative). We say that $L$ is a separable algebraic extension iff every element of $L$ is separable algebraic.

These definitions, of course, are quite standard. Now what I'd like to find is a proof of such standard facts as "if $x$ is algebraic separable over $K$ then $K(x)$ is separable" and "if $L$ is algebraic separable over $K$ and $M$ is such over $L$ then $M$ is such over $K$", or a definition of the separable degree of an extension, all without using field automorphisms or the trick of counting embeddings in an algebraic closure.

(There can be a number of reasons to want this: for pedagogical purposesout of a desire to postpone a discussion of Galois theory at a later point, or because embeddings/automorphisms are computationally or logically more complex objects than field extensions, or simply because it seems that the point of view in the first paragraph above should be more natural, or to compare different points of view.)

Now every textbook I could find on field extensions uses at some point a comparison between the number of embeddings of $L$ in the algebraic closure of $K$ and the degree $[L:K]$. But surely this can be avoided (we can, instead, work explicitly with roots of polynomials and perhaps elementary symmetric functions).

So, does someone know a place where separable field extensions are introduced without counting embeddings or similar objects, staying as close as possible to the definition I gave above?

Edit: Maybe the nicest definition of an algebraic $x$ being separable over $K$ of characteristic $p$ is that $K(x) = K(x^p)$.

• I have a feeling this can be done by resultants. If $y$ is in $K(x)$, then $y=g(x)$ for some polynomial $g$ with coefficients in $K$, and the minimal polynomial $Q(X)$ for $y$ is a factor of ${\rm Res}_t(X-g(t),f(t))$. – Gerry Myerson Feb 27 '16 at 0:38

After giving this some thought, here's how it might work:

• Define an element $x$ of an extension field of $k$ to be algebraic separable over $k$ iff it is algebraic and its minimal polynomial $f$ over $k$ is relatively prime with $f'$, which is equivalent to saying $f' \neq 0$. Since any polynomial $f$ in characteristic $p>0$ can be written uniquely as $f(t) = f_0(t^{p^e})$ for some $e$ with $f_0 \neq 0$, in the context where $x$ has $f$ as minimal polynomial, $f_0$ is irreducible and $x$ is irreducible iff $e=0$.

• Proposition 1: over a field $k$ of characteristic $p>0$, if $f(t) = f_0(t^{p^e})$ with $e>0$ and $f_0$ monic, then $f$ is reducible iff the coefficients of $f_0$ (equivalently, those of $f$) are $p$-th powers, and in this case $f$ is, in fact, a $p$-th power. The "if" part is easy, and the "only if" part can be proved by reducing to $e=1$, considering the factorization of $f_0^p$ inside $k^p[t]$ and using the following lemma: if $h \in k[t]$ satisfies $h^i \in k^p[t]$ for some $1\leq i<p$ then in fact $h \in k^p[t]$ (this can be seen using a Bézout relation $u i = 1 + v p$ with $u,v\in\mathbb{N}$).

• Proposition 2: if $x$ is algebraic over $k$ of characteristic $p>0$ then exactly one of the following statement holds: either (a) $x$ is separable, the minimal polynomial of $x^p$ over $k$ has coefficients in $k^p$ and $\deg(x) = \deg(x^p)$ and $k(x) = k(x^p)$, or (b) $x$ is not separable, the minimal polynomial of $x^p$ over $k$ does not have all its coefficients in $k^p$ and $\deg(x) = p\cdot\deg(x)$. (This is easy using proposition 1.)

• Proposition 3: if $k \subseteq K$ is a finite extension of fields of characteristic $p>0$ and $K^p$ spans $K$ as a $k$-vector space, then $K$ is separable over $k$ (meaning that every element of $K$ is separable over $k$). This is basically saying that the extensions $K^p$ and $k$ of $k^p$ (inside $K$) are linearly disjoint, but we don't really need all the machinery of linear disjointness to prove it. Proof. Let $x_1,\ldots,x_d$ be a basis of $K$ as a $k$-vector space (where $d = [K:k]$) and let $y \in K$ have degree $d'$: write $y^j = \sum_{i=0}^{d-1} c_{i,j} x_i$ for $0\leq j\leq d'-1$ on the chosen basis: since $1,y,\ldots,y^{d'-1}$ are $k$-linearly independent, the matrix $c_{i,j}$ has rank $d'$; but raising to the $p$-th power, we have $y^{pj} = \sum_{i=0}^{d-1} c_{i,j}^p x_i^p$. Now the hypothesis that $K^p$ spans $K$ as a $k$-vector space implies that $x_1^p,\ldots,x_d^p$ do so, so they are a basis of $K$ as a $k$-vector space. And the matrix of the $c_{i,j}^p$ has the same rank as that of the $c_{i,j}$ since Frobenius is an isomorphism from $k$ to $k^p$, and the rank of a matrix does not depend on the field where it is computed. So from $y^{pj} = \sum_{i=0}^{d-1} c_{i,j}^p x_i^p$ we deduce that $1,y^p,\ldots,y^{p(d'-1)}$ are linearly independent over $k$, that is, $y^p$ has degree $d' = \deg(y)$, and by proposition 2 that $y$ is separable. End of proof.

• Proposition 4: if $x_1,\ldots,x_n$ are such that $x_i$ is algebraic separable over $k(x_1,\ldots,x_{i-1})$, then $k(x_1,\ldots,x_n)$ is (algebraic) separable over $k$. Proof: in characteristic $0$ there is nothing to prove, and in characteristic $p>0$ we have $k(x_1) = k(x_1^p)$ because $x_1$ is separable over $k$, then $k(x_1)(x_2) = k(x_1)(x_2^p) = k(x_1^p)(x_2^p)$ because $x_2$ is separable over $k(x_1)$, and so on; so $k(x_1,\ldots,x_n) = k(x_1^p,\ldots,x_n^p)$ so it is easy to see that the monomials in $x_1^p,\ldots,x_n^p$ span $k(x_1,\ldots,x_n)$ as a $k$-vector space, so proposition 3 applies.

• The following statements are then almost trivial: if $(x_i)_{i\in I}$ are all algebraic separable over $k$ then the extension $k(x_i)_{i\in I}$ they generate is (algebraic) separable. (Proof: it is enough to prove it for a finite subfamily, a case which is contained in proposition 4.) And if $k \subseteq K \subseteq L$ is a tower of fields with $K$ algebraic separable over $k$ and $L$ algebraic separable over $K$ then $L$ is (algebraic) separable over $k$. (Proof: take $y \in L$ and $x_1,\ldots,x_n \in K$ the coefficients of its minimal polynomial over $k$; then proposition 4 applies to $k(x_1,\ldots,x_n,y)$.)

• It then makes sense to define the relative (algebraic) separable closure of $k$ in some extension field $K$ as the extension generated by all elements of $K$ which are algebraic separable over $k$, and which is, in fact, the set of all such elements by the above. We can say that $K$ is a purely inseparable extension of $k$ when $k$ is equal to its separable closure in $K$, and this is equivalent to the minimal polynomial over $k$ of every element of $K$ to being of the form $t^{p^e} - c$ for some $c \in k$.

I didn't work out the properties of the separable degree in as much detail, but we can define $[K:k]_{\mathrm{sep}}$ it as the degree of the relative (algebraic) separable closure of $k$ inside $K$, which makes sense because of the above. The crucial point to showing that the separable degree is multiplicative is to show that if $k \subseteq K$ is purely inseparable and $K \subseteq K'$ is finite separable and $k'$ the separable closure of $k$ in $K'$ then $[k':k] = [K':K]$, meaning essentially that $K$ and $k'$ are linearly disjoint extensions of $k$ inside $K'$: again, the machinery of linear disjointness can certainly be avoided.

To summarize, I think it's best to think in terms of linear disjointness ("MacLane's criterion"), if not explicitly at least implicitly.

I'm still interested in knowing whether any textbook uses this approach.