Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 18I (Galois Theory)

A finite extension $L/K$ is normal when every irreducible polynomial in $K[T]$ having one root in $L$ splits completely over $L$.

Suppose first that $L/K$ is normal. Choose ${\alpha }_1,\dots ,{\alpha }_m$ with $L=K({\alpha }_1,\dots ,{\alpha }_m)$, and let $f_j$ be the minimal polynomial of ${\alpha }_j$ over $K$. Every $f_j$ splits in $L$ by normality, and $L$ is generated by their roots because it is already generated by the particular roots ${\alpha }_j$. Thus $L$ is the splitting field of $f_1\cdots f_m$.

Conversely, suppose that $L$ is the splitting field of $f\in K[T]$. Every $K$-embedding $\sigma :L\to \overline{K}$ permutes the roots of $f$, so $\sigma (L)=L$. By the embedding characterization of normality, $L/K$ is normal. This proves the finite normal extension as a splitting field criterion.

True. If $M/K$ is normal, the finite normal extension as a splitting field criterion writes $M$ as the splitting field over $K$ of some $f\in K[T]$. Since $K\subseteq L\subseteq M$, the same field $M$ is the splitting field of $f$, now regarded as a polynomial over $L$. Hence $M/L$ is normal.

False. Letwhere $\omega$ is a primitive cube root of unity. Then $M$ is the splitting field of $T^3-2$ over $\mathbb{Q}$, so $M/K$ is normal. The irreducible polynomial $T^3-2$ has a root in $L$ but does not split there, so $L/K$ is not normal.

False. TakeEach of $L/K$ and $M/L$ is a quadratic extension and hence normal. However, $T^4-2$ has the root $\sqrt[4]{2}$ in $M$ while its roots $i\sqrt[4]{2}$ and $-i\sqrt[4]{2}$ are absent. Thus $M/K$ is not normal.

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.