# Abstract Nonsense

## Separable Extensions (Pt. I)

Point of Post: In this post we discuss separable extensions and the separable closure of an extension.

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Motivation

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Last time we discussed normal extensions which were the nice kind of extensions which (among several other definitions) are just the splitting field for a set of polynomials over the ground field. In this post we’d like to discuss another kind of “nice” extension, namely separable extensions. Roughly a separable extension is one where there are multiple roots which become indistinguishable (inseparable, if you will) to the Galois group of the extension. For example, consider the finite field $\mathbb{F}_p$ and the rational function field $\mathbb{F}_p(t)$. Then, we claim that $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$ is a simple extension of degree $p$. Indeed, it’s a simple extension since $\mathbb{F}_p(t)=\mathbb{F}_p(t^p)(t)$. To see that it’s degree $p$ it suffices to show that $u^p-t^p\in\mathbb{F}_p(t)[u]$ is irreducible (since it annhilates $t$). To do this we merely note that if $u^p-t^p$ factors in $\mathbb{F}_p(t^p)[u]$ it factors in $\mathbb{F}_p(t)[u]$ but we already know the irreducible factorization there–$u^p-t^p=(u-t)^p$. Clearly then (since everything in sight if a UFD) the factorization in $\mathbb{F}_p(t)[u]$ must look like $(u-t)^m(u-t)^n$ for some $m,n$. But this, in particular, implies that $t^m\in\mathbb{F}(t^p)$ which clearly says that $m=p$ and thus our factorization was trivial.  Ok, so what? The important thing to note is that while $[\mathbb{F}_p(t):\mathbb{F}_p(t^p)]=p$ you can easily verify that $\text{Gal}(\mathbb{F}_p(t)/\mathbb{F}_p(t^p))$ is trivial since they are just permutations of the roots of $u^p-t^p=(u-t)^p\in\mathbb{F}_p(t)[u]$ [which, in case this isn’t obvious, there is only one such root].

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Similar to the case of $\mathbb{Q}(\sqrt[3]{2})$ we see that the Galois group of our extension has order strictly less than the degree of our extension. The key observation though is that it is for a fundamentally different reason. The reason that $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})|<[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ is an issue of normality [in the sense of normal extensions]–the polynomial, for which the Galois group is permuting its roots, doesn’t have all of its root in the field. In our case though the polynomial the Galois group is permuting roots of has all of its roots in the field, the problem is that it has repeated roots. The Galois group’s raison d’etre is to permute the roots the best it can, but when there are repeated roots the Galois group has a difficulty doing this because it can’t distinguish (separate) the repeated roots.

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Thus, our topic of this thread will be those fields which do not impede the Galois group in doing its job, or at least do not impede it in the way described above.

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Separable Extensions

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We begin with the basic objects of study–separable polynomials. We say that a polynomial $f(x)\in F[x]$ is separable if it has distinct roots in $\overline{F}$ (the algebraic closure of $F$). For example, $x^2-2\in\mathbb{Q}[x]$ is separable since it has distinct roots $\pm\sqrt{2}$ in $\overline{\mathbb{Q}}\subseteq\mathbb{C}$. If a polynomial is not separable we call it inseparable.

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As defined it’s not totally obvious how to figure out if a given polynomial is separable. Lucky for us though there is a very simple characterization of separable polynomials.

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Theorem: Let $f(x)\in F[x]$. Then, $f(x)$ is separable if and only if $(f,f')=1$.

Proof: Suppose that $f(x)=(x-\alpha_1)\cdots(x-\alpha_n)\in\overline{F}[x]$. Then, $\displaystyle f'(x)=(x-\alpha_2)\cdots(x-\alpha_n)+\cdots+(x-\alpha_1)\cdots(x-\alpha_{n-1})$. Suppose first that $f$ is separable, then evidently each $\alpha_1,\cdots,\alpha_n$ are distinct and so clearly each $x-\alpha_i$ appears in all but one term in the expression for $f'(x)$ and so clearly the two are coprime in $\overline{F}[x]$ and so clearly coprime in $F[x]$ Conversely, suppose that $(f,f')=1$. Then, we know we can write

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For example, if $F$ is characteristic $p$ then $x^p-a$ is not separable since $(x^p-a)'=px^{p-1}=0$ (since $p=0$).

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In fact, this allows us to conclude the following:

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Theorem: Let $f(x)\in F[x]$ be irreducible. Then, $f$ is separable if and only if $f'\ne 0$. In particular, if $F$ is characteristic zero then irreducibles are always separable and if $F$ is characteristic $p$ then $f$ is inseparable if and only if $f=g(x^p)$ for some $g(x)\in F[x]$

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We know define the notion of a separable extension. We call a finite extension $k/F$ separable if $m_{\alpha,F}\in F[x]$ is separable for all $\alpha\in k$. We call an element of the extension separable if its minimal polynomial is separable–and thus we see that an extension is separable if and only if all of its members are separable. We obviously define an inseparable extension to be a non-separable extension.

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For example, any finite extension of $\mathbb{Q}$ (or, in fact, any characteristic zero field) is separable since minimal polynomials are irreducible, and irreducibles are separable.

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The extension $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$ is not separable since, as we showed in the motivation, $m_{t,\mathbb{F}_p(t^p)}=u^p-t^p\in \mathbb{F}_p(t^p)[u]$ which is inseparable since it’s derivative (with respect to $u$!) is zero [or, it’s a polynomial in $u^p$].

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Now, the first thing I’d like to do is show that a separable extension $k/F$ always has $[k:F]$ embeddings into some extension $K/F$. We then use this to show that first non-trivial fact about extensions–the primitive element theorem. The basic idea behind this is that if, for example, $F(\alpha)$ is separable, so that $m_{\alpha,F}$ is separable, we can find $\deg m_{\alpha,F}=[F(\alpha):F]$ embeddings of $F(\alpha)$ into a splitting field of $m_{\alpha,F}$ by taking $\alpha$ to each of the other roots of $m_{\alpha,F}$. The idea isn’t quite as simple since we only know that a finite separable extensions looks like $F(\alpha_1,\cdots,\alpha_n)$ but the idea is still there:

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Theorem(Primitive Element Theorem): Let $k/F$ be a finite separable extension of degree $n$ and let $\sigma:k\to K$ be some embedding of fields. Then, there exists an extension $L/K$ such that the number of extensions of $\sigma$ to an embedding $k\to L$ is $n$.

Proof: We know that $k=F(\alpha_1,\cdots,\alpha_r)$ for some separable (since everything is separable!) $\alpha_1,\cdots,\alpha_r\in k$. Let $L$ be a splitting field for $\sigma(m_{\alpha_1,F}),\cdots,\sigma(m_{\alpha_r,F})$ over $K$. We will show that the number of embeddings $k\to L$ extending $\sigma$ is $n$. To do this we shall induct on $[L:K]$. If $[L:K]=1$ we’re done, so assume that it’s true for all extensions with $[L:K] and let $[L:K]=N$. Clearly at least one of the $\alpha_i$‘s is not in $F$ (otherwise $[L:K]=1$), we assume that $i=1$. Now, it’s obvious that the number of extensions of $\sigma$ with $F(\alpha_1)\to L$ is just the number of roots of $\sigma(m_{\alpha_1,F})$ in $L$ (this is just the isomorphism extension theorem). Consider now $k$ as $F(\alpha_1)(\alpha_2,\cdots,\alpha_r)$. Note that since $[L:\sigma(F(\alpha_1))] we know, by the inductino hypothesis,  that the number of extensions of any given extension $\tau:F(\alpha_1)\to L$ is $[k:F(\alpha_1)]$. Since the number of extensions of $\sigma$ to $F(\alpha_1)\to L$ is $[F(\alpha_1):F]$ we may conclude that the number of extensions of $\sigma$ to $k\to L$ is $[k:F(\alpha_1)][F(\alpha_1):F]=n$. $\blacksquare$

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.print