Here is the question I started with.

Give an example of an algebraic extension $L$ of a field $k$ of positive characteristic such that $\left|\text{Aut}_kL\right|<\infty$ but $[L:k]$ is not.

Since the fundamental theorem of Galois theory gives a one-one correspondence between extension dimensions and automorphisms, clearly the extension should not be a Galois extension.

So, initially the $\left| \text{Aut}_kL \right|<\infty$ condition tempted me to go along the lines of taking $k=\mathbb{Q}$ and $L=\mathbb{Q}\left(2^{1/3},2^{1/9},2^{1/27}, \cdots \right)$ so that $[L:k]=\infty$ and the automorphism group would be trivial. (I did not take $\left(2^{1/2}, 2^{1/4},2^{1/8},\cdots \right)$ just because I wanted to arrive at the trivial automorphism group.) But of course, this is a $\underline{\text{characteristic zero example}}$ and works because the extension I mentioned is not normal, hence not Galois.

Next, keeping $\text{char}(k)>0$ in mind, I tried working with $k=\mathbb{F}_p$ and $L=\mathbb{F}_p(t)$ with $t$ transcendental over $\mathbb{F}_p(t)$, honestly $\underline{\text{forgetting that }} L \underline{\text{ has to be algebraic}}$, and turned up with the fact that $\sigma \in \text{Aut}_kL \implies \sigma(t)=\dfrac{at+b}{ct+d}, \ ac\ne 0 \ne ad-bc \quad \quad$ from the well-known fact that $[k(t):k(u)]=\max(\deg g, \deg f)$ where $u=\dfrac{f(t)}{g(t)}$, $f,g \ \text{coprime}, \ f,g \in k[t]\backslash k $, $\quad \quad$ which actually gave $\left|\text{Aut}_kL\right|\le p^3(p-1)<\infty$ and $[L:k]=\infty$.

Realizing my mistake, I tried to work the characteristic zero example into a field of characteristic $p>0$, which led me to think of a purely inseparable extension $L/k$, since in such an extension, there will be no question of conjugacy, so maybe that will give me a trivial automorphism group.

So, I took $k$ to be a field of characteristic $p>0$ such that $a\in k\backslash k^p \ne \phi$, where $k^p=\{x^p \mid x\in k\}$ and $L$ to be the purely inseparable closure of $k$ in $\bar{k}$ [the intuition being the want to liken it to $k(a^{1/p},a^{1/{p^2}},a^{1/{p^3}},\cdots)$].
For the proof of $[L:k]=\infty$, I assumed it to be finite, say $[L:k]=n$ and for $m>n$, took the ${p^m}$-th root of $a$, $\beta$ in $\bar{k}$, and considered the minimal polynomial of $\beta$ over $k$. Showing $\left(X^{p^m}-a\right)$ to be the minimal polynomial of $\beta$ (by contradicting $a\notin k^p$), we could contradict the assumption that $[L:k]$ is bounded.
That $\text{Aut}_kL$ is finite is trivial from the fact that the minimal polynomial of any $\alpha \in L$ has to be $(X-\alpha)^{p^l}$ for some $l\ge 0$, so $\sigma \in \text{Aut}_kL \implies \sigma(\alpha)=\alpha\implies $ the automorphism group must be trivial.
Finally, to show that such a field, that is, a non-perfect field of positive characteristic exists at all, $\mathbb{F}_p(t)$ served as an example.

So here is my question: I took $L=$ the purely inseparable closure of $k$ in $\bar{k}$, and not $L=k\left(a^{1/p}, a^{1/{p^2}},a^{1/{p^3}},\cdots \right)$ just because the proof of $\left| \text{Aut}_kL \right| <\infty$ seemed imminent from the former but essentially, in my proof I do not seem to have used much from the larger choice of $L$ other than that one $a$ that is not a $p^{th}$ power. Please tell me if my proof seems wrong. If it is not wrong and there is a different choice for $L$ or someone can prove it with the latter choice here, I would be very happy to see it. Along the way, it'd be awesome if you can provide me with any intuition in the role that perfect fields and purely inseparable extensions play in Galois theory (which I'm only just learning) as a takeaway. Thanks in advance.

  • $\begingroup$ @JyrkiLahtonen I was discussing this issue with a friend before I worked out the solution and even though the question was framed in that way, with a heading of "In the following, $k$ is a field of positive characteristic", I am sure it actually asked to exemplify a pair of fields $k,L$ and not only $L$, because we remembered the result that all finite fields are perfect. I have edited the question to avoid confusions in that regard. $\endgroup$ Commented Jul 7, 2020 at 5:03
  • $\begingroup$ Ok. That makes sense. And your construction looks good to me. $\endgroup$ Commented Jul 7, 2020 at 5:06
  • $\begingroup$ Thank you so much for confirming, I was afraid it was going too much in circles. :D $\endgroup$ Commented Jul 7, 2020 at 5:25


You must log in to answer this question.

Browse other questions tagged .