# Abstract Nonsense

## Separable Extensions (Pt. II)

Point of Post: This is a continuation of this post.

$\text{ }$

From this we can prove the, extremely useful, primitive root theorem. Recall that simple extension is a singly generated one:

$\text{ }$

Theorem: Let $k/F$ be a finite separable extension, then $k$ is simple.

Proof: If $F$ is finite then $k$ is finite and the conclusion follows from the cyclicity of $k^\times$.

$\text{ }$

Suppose now that $F$ is infinite, so that $k$ is infinite. Write $k$ as $F(\alpha_1,\cdots,\alpha_n)$ where each $\alpha_i$ is separable. It suffices to show that any separable extension of the form $k=F(\alpha,\beta)$ is simple since the general case then follows by induction. So, let $L$ be the extension of $F$, as constructed in the previous theorem, with $\text{id}:F\to F$ being the $\sigma$ map. We then know that there are $[k:F]$ extensions of $\text{id}$ to a map $k\to L$. Let $\xi\in F$ and consider $F(\alpha+\xi\beta)\leqslant k$. If $[F(\alpha+\xi\beta):F]<[k:F]$ then we can find two different extensions $\tau,\tau'$ of the identity map on $F$ to a map $k\to L$ which are equal on $F(\alpha+\xi\beta)$ (why?). Note then that evidently $\tau(\beta)\ne\tau'(\beta)$ otherwise the equation $\tau(\alpha)+\xi\tau(\beta)=\tau'(\alpha)+\xi\tau'(\beta)$ would imply that $\tau(\alpha)=\tau'(\alpha)$ and thus $\tau=\tau'$ on $k$ (contradictory to assumption). Thus, we see that $\displaystyle \xi=\frac{\tau(\alpha)-\tau'(\alpha)}{\tau(\beta)-\tau'(\beta)}$. Now, since there are only finitely many $\tau,\tau'$ this clearly implies that there are only finitely many $\xi\in F$ with $[F(\alpha+\xi\beta):F]<[k:F]$ and so, since $F$ is infinite, there exists $\xi_0\in F$ with $F(\alpha+\xi_0\beta)=k$ as desired. $\blacksquare$

$\text{ }$

Since any extension of a characteristic zero field is separable we get the following corollary:

$\text{ }$

Corollary: Every finite extension of a characteristic zero field is simple.

$\text{ }$

In particular, I feel compelled to state the obvious:

$\text{ }$

Corollary: Every finite extension of $\mathbb{Q}$ is simple.

Now that we have the primitive element we can state what is, perhaps, the most functional definition of separability of extensions. Just to note this before hand, it involves the tensor products of algebras. Namely:

$\text{ }$

Theorem: Let $k/F$ be a finite extension. Then, $k/F$ is separable if and only if the $F$-algebra $k\otimes_F \overline{F}$ contains no nonzero nilpotents.

Proof: Suppose first that $k$ is separable. By the primitive element theorem we can write $k$ as $F(\gamma)$ for some separable $\gamma$. It is well-known then that we have an $F$-map $\overline{F}[x]\to F[x]/(m_\gamma)\otimes \overline{F}$ defined by

$\text{ }$

$\displaystyle \sum_i a_ix^i\mapsto \sum\left[(x_i+(m_\gamma))\otimes a_i\right]$

$\text{ }$

with kernel $(m_\gamma)$.  But, this is actually a map of $F$-algebras and thus we may conclude that, as $F$-algebras,

$\text{ }$

$\overline{F}[x]/(m_\gamma)\cong (F[x]/(m_\gamma))\otimes_F \overline{F}\cong F(\gamma)\otimes_F \overline{F}$

$\text{ }$

But, since $m_\alpha(\gamma)$ splits into distinct linear factors we may conclude then by the Chinese remainder theorem that $F(\gamma)\otimes_F \overline{F}\cong \overline{F}^{n}$ where $n=[k:F]$–and since the product of fields never has any nonzero nilpotents we may conclude that our ring has no non-zero nilpotents.

$\text{ }$

Conversely, suppose that $k$ is not separable. There then exists some non-separable $\alpha\in k$. Note then that since the inclusion $F(\alpha)\hookrightarrow k$ gets tensored to an inclusion $F(\alpha)\otimes_F \overline{F}\hookrightarrow k\otimes_F\overline{F}$ (since all modules over vector spaces are free, and free modules are flat), and thus if we can show that $F(\alpha)\otimes_F\overline{F}$ has a nonzero nilpotent so will $k\otimes_F\overline{F}$. That said, the above analysis shows that, once again, $F(\alpha)\otimes_F\overline{F}$ is just $\overline{F}/(m_\alpha)$ as $F$-algebras. But, now we know that $m_\alpha=(x-\beta)^2g(x)$ for some polynomial $g(x)\in\overline{F}[x]$. We obviously see then that $(x-\beta)g(x)+(m_\alpha)$ is non-zero but its square $m_\alpha g(x)$ is zero. The conclusion follows. $\blacksquare$

$\text{ }$

With this we can easily prove the two following theorems:

$\text{ }$

Theorem: Let $F(\alpha_1,\cdots,\alpha_n)/F$ be a finite extension. Then, the extension is separable if and only if each $\alpha_1,\cdots,\alpha_n$ is separable.

Proof: We clearly get the if way for free. Conversely,it suffices to show that $F(\alpha_1,\cdots,\alpha_n)\otimes_F\overline{F}$ contains no nonzero nilpotents. What we claim is that $F(\alpha_1,\cdots,\alpha_n)\cong\overline{K}^{[F(\alpha_1,\cdots,\alpha_n):F]}$–we prove this by induction. By the previous theorem we can see in particular that if $K(\alpha)/K$ is any field extension with $K(\alpha)$ separable then $K(\alpha)\otimes_K \overline{K}=\overline{K}^{[K(\alpha):K]}$ which gives us the base case. For the inductive step we note that

$\text{ }$

\begin{aligned}F(\alpha_1,\cdots,\alpha_n)\otimes_F\overline{F} &\cong F(\alpha_1,\cdots,\alpha_n)\otimes_{F(\alpha_1)}(F(\alpha_1)\otimes\overline{F})\\ &\cong F(\alpha_1,\cdots,\alpha_n)\otimes_{F(\alpha_1)}\overline{K}^{[F(\alpha_1):F]}\\ & \cong \left(F(\alpha_1)(\alpha_2,\cdots,\alpha_n)\otimes_{F(\alpha_1)}\overline{K}\right)^{[F(\alpha_1):F]}\end{aligned}

$\text{ }$

but by inductino hypothesis $F(\alpha_1)(\alpha_2,\cdots,\alpha_n)\otimes_{F(\alpha_1)}\overline{K}\cong \overline{K}^{[F(\alpha_1,\cdots,\alpha_n):F(\alpha_2,\cdots,\alpha_n)]}$. Thus,

$\text{ }$

\begin{aligned}F(\alpha_1,\cdots,\alpha_n)\otimes_F\overline{K} & \cong\left(F(\alpha_1)(\alpha_2,\cdots,\alpha_n)\otimes_{F(\alpha_1)}\overline{K}\right)^{[F(\alpha_1):F]}\\ &\cong \overline{K}^{[F(\alpha_1,\cdots,\alpha_n):F(\alpha_2,\cdots,\alpha_n][F(\alpha_1):F]}\\ &=\overline{K}^{[F(\alpha_1,\cdots,\alpha_n):F]}\end{aligned}

$\text{ }$

The theorem then follows since products of fields never have any nonzero nilpotents. $\blacksquare$

$\text{ }$

Theorem: Let $L/k/F$ be a tower of finite fields. Then, $L/F$ is separable if and only if $L/k$ and $k/F$ is separable.

Proof: We have that if $L/k$ and $k/F$ are separable

$\text{ }$

$L\otimes_F\overline{F}\cong L\otimes_k(k\otimes_F\overline{F})\cong L\otimes_k\overline{F}^{[k:F]}\cong (L\otimes_k\overline{F})^{[k:F]}\cong (\overline{F}^{[L:k]})^{[k:F]}\cong \overline{F}^{[L:F]}$

$\text{ }$

where we made use of the fact that since $k/F$ is algebraic that $\overline{k}=\overline{F}$. We may clearly then conclude that $L\otimes_F\overline{F}$ has no nonzero nilpotents and thus $L/F$ is separable.

$\text{ }$

Conversely, suppose that $L/F$ is separable. The inclusion $k\otimes_F\overline{F}\hookrightarrow L\otimes_F\overline{F}$ implies that $k\otimes_F\overline{F}$ has no nilpotents and so $k/F$ is separable. The fact that $L/K$ is separable follows from the fact that $m_{\alpha,K}\mid m_{\alpha,F}$ in a splitting field. $\blacksquare$

$\text{ }$

$\text{ }$

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