\(\newcommand{\Aut}{\operatorname{Aut}}\)\(\newcommand{\from}{\colon}\)\(\newcommand{\cross}{\times}\)\(\newcommand{\cover}[1]{{\widetilde{#1}}}\)\(\newcommand{\CC}{\mathbb{C}}\)\(\newcommand{\FF}{\mathbb{F}}\)\(\newcommand{\QQ}{\mathbb{Q}}\)\(\newcommand{\RR}{\mathbb{R}}\)\(\newcommand{\ZZ}{\mathbb{Z}}\)Questions handed in by students on 2018-11-02.

Exercises -

1. Isn't \(x^n + x + p\) irreducible for all \(p\) prime rather than
just three?  A Hint for the proof?

Answer: If \(p = 2\) then the polynomial factors for some \(n\).

Hint: Compute the complex roots of the polynomial numerically (and to
a fairly high precision).  Plot the roots in \(\CC\) and stare at the
resulting picture for a goodly time.

Lecture -

1. In Lemma 58 isn't this just linear independence of maps in \( \{ f
\from L \to L \mid \mbox{$f$ is linear}\} \)?

Answer: Yes.  However, I want to very carefully distinguish between
automorphisms (which are always linear maps) and linear maps (which
are often not automorphisms).

2. is there a vector space that contains automorphisms as elements?
Either abstractly or constructively.

Answer: Yes.  We can be very concrete here, if we like.  Suppose that
\(L\) is a vector space over \(K\).  Choose a basis.  This identifies
the set of \(K\)-linear maps from \(L\) to itself with the set \(M\)
of (correctly sized, square) matrices.  Now, \(M\) is itself a vector
space; inside of \(M\) we can find a "copy" of the automorphism group.

3. In proof of 54: Why does at most \(d\) possibilities for
\(\alpha_i\) and \(d\) poss. for \(\sigma(\alpha_i)\) \( \implies \)
\(|\Aut(L/K)| \leq d^2 \)? (i.e. the final step of 54's proof).

Answer: How embarrassing!  I said and wrote \(d^2\).  But the correct
upper bound is \(d^d\); there are \(d\) choices and each choice is
among \(d\) things.

4. I can see why \( x^p - 1 = (x - 1)^p \).  But why is \( x^p - t =
(x - \alpha)^p \). (in \( \FF_p[x]/(x^p - t) \) example)

Answer: The proofs are the same; use the binomial theorem to expand
the product and notice that all terms except for the first and last
are divisible by \(p\) and thus are zero.

Terminology -

1. \( \varphi \) This is pronounced Fye not Fee.  As in the "Fi" of
Five.

Answer:
https://english.stackexchange.com/questions/11363/why-are-greek-letters-pronounced-incorrectly-in-scientific-english