Question on notes rings of integers without power basis from KConrad I am currently reading the following excerpt from a note of Keith Conrad on algebraic number theory. However, I have some questions because of the gaps in my understanding.

The unique cubic subfield $K_p$ corresponds under the Galois correspondence to the unique subgroup $H$ of index 3 in $\mathbf{F}_p^\times$. We have the map
$$f:\mathbf{F}_p^\times\to \mathbf{F}_p^\times,x\mapsto x^3$$
which has image the subgroup of all cubes in $\mathbf{F}_p^\times$ and kernel the unique subgroup $S$ of order 3. Therefore by the first isomorphism theorem, $\mathbf{F}_p^\times/S\cong \{\text{cubes in }\mathbf{F}_p^\times\}$. Therefore $H=\{\text{cubes in }\mathbf{F}_p^\times \}$. The Galois group of $K_p$ over $\mathbf{Q}$ is then the quotient $\mathbf{F}_p^\times/H$. Is there a shorter way to see this?
By "splits completely in $K_p$, I think he means in $\mathcal{O}_{K_p}$ (the ring of integers of $K_p$), right?
However, I don't understand the sentence "$q$ splits completely in $K_p$ if and only if its Frobenius in $\operatorname{Gal}(K_p/\mathbf{Q})$ is trivial, which is equivalent to $q$ being a cube modulo $p$." What is this "Frobenius" ?

.. so $f$ splits completely in $(\mathbf{Z}/2\mathbf{Z})[X]$. Why is this the case?
 A: 
Is there a shorter way to see this?

Because $p\equiv1\pmod{3}$ the Galois group, which is isomorphic to $\Bbb{F}_p^{\times}$, has a unique subgroup $\Bbb{F}_p^{\times3}$ of index $3$. By the Galois correspondence this means $\Bbb{Q}(\zeta_p)$ has a unique subfield $K_p$ of degree $3$ over $\Bbb{Q}$. This is how the author defines $K_p$, and because the subgroup is normal the extension $K_p/\Bbb{Q}$ is Galois. The map
$$\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q})\ \longrightarrow\ \operatorname{Gal}(K_p/\Bbb{Q}):\ \sigma\ \longmapsto\ \sigma\vert_{K_p},$$
is surjective with kernel $\Bbb{F}_p^{\times3}$, so by the first isomorphism theorem
$$ \operatorname{Gal}(K_p/\Bbb{Q})\cong\Bbb{F}_p^{\times}/\Bbb{F}_p^{\times3}.$$

By "splits completely in $K_p$, I think he means in $\mathcal{O}_{K_p}$ (the ring of integers of $K_p$), right?

That's right. When discussing primes of a number field $K$, one always means the prime ideals of $\mathcal{O}_K$, unless stated otherwise explicitly.

What is this "Frobenius" ?

The Galois group $\operatorname{Gal}(K-p/\mathbb{Q})$ acts on the set of primes lying over $q$. For every prime $\mathfrak{q}$ lying over $q$, the decomposition group $G_{\mathfrak{q}}$ is the stabilizer of $\mathfrak{q}$, and the residue field of $\mathfrak{q}$ is the field $k_{\mathfrak{q}}=\mathcal{O}_{K_p}/\mathfrak{q}$. Becaus the Galois group acts transitively on the set of primes over $q$, all decomposition groups are isomorphic (and even conjugate in the Galois group), and all residue fields are isomorphic. Moreover the natural map
$$G_{\mathfrak{q}}\ \longrightarrow\ \operatorname{Gal}(k_{\mathfrak{q}}/\Bbb{F}_q),$$
is surjective. Now the Frobenius automorphism of $k_{\mathfrak{q}}$ is the map
$$k_{\mathfrak{q}}\ \longrightarrow\ k_{\mathfrak{q}}:\ x\ \longmapsto\ x^q,$$
which trivial if and only if $k_{\mathfrak{q}}=\Bbb{F}_q$, if and only if $k_{\mathfrak{q}}=\Bbb{F}_q$ for every prime $\mathfrak{q}$ lying over $q$, which is equivalent to $q$ splitting completely in $K_p$.
The different Frobenius automorphisms for the different primes lying over $q$ together lift to a $q$-Frobenius map on $K_p$, wich is trivial if and only if the Frobenius automorphisms are trivial.
But quite frankly, if you haven't heard the term 'Frobenius' at all, you should be reading up on this first. I believe Keith Conrad's notes are usually quite complete and well-organised, so there is likely to be ample explanation of all the relevant theory earlier in the notes, or in earlier notes.

.. so $f$ splits completely in $(\mathbf{Z}/2\mathbf{Z})[X]$. Why is this the case?

In essence, it is the observation that
$$\Bbb{F}_2[X]/(f)\cong\Bbb{Z}[X]/(2,f)\cong\Bbb{Z}[\alpha]/(2).$$
Can you deduce from here that $f\subset\Bbb{F}_2[X]$ splits completely because $(2)\subset\Bbb{Z}[\alpha]$ splits completely? I suggest you read up on  the Kummer-Dedekind theorem, which generalizes this idea and makes it more precise. It is an absolutely fundamental result in algebraic number theory.
