# Is there a finite field extension with „less“ intermediate fields than Aut-Subgroups?

When looking for a counterexample to the „lattice isomorphism“ part of the fundamental theorem of Galois theory, I usually look for non-Galois extensions $$K\colon F$$ e.g. by choosing $$F$$ to not contain all conjugates of some element. Consider for example $$\mathbb Q(\sqrt{2})\colon \mathbb Q$$ which has trivial automorphism group, but a two-element lattice of intermediate fields.

Are there examples (in char. 0, since I don't understand anything else) where we have more subgroups of $$\operatorname{Aut}(K\vert F)$$ than there are intermediate fields?

…Or is the map $$\varphi\colon \operatorname{Sub}(\operatorname{Aut}(K\vert F)) \to \{E\mid F\leq E \leq K\} \\ H \mapsto \operatorname{Fix}(H)$$ injective even for non-Galois (but still finite) extensions?

• Jul 3, 2019 at 15:06
• In hopes that the downvoter reads that: Care to explain the downvote? Is there a “commonly known” answer, or was something else about this question substandard? I try to be very clean. Jul 3, 2019 at 17:51
• A Galois extension $K/E$ is such that $E = K^G$ (the subfield fixed by $G$) for some finite subgroup $G \le Aut(K)$. For an intermediate field $K/L/K^G$ any $K^G$-monomorphism $L \to K$ extends to an automorphism of the normal closure $K$. Thus $Aut(K/K^G) / Aut(K/L)$ is the set of $K^G$-embeddings $L \to K$. In other words to each subgroup of $G$ there is an intermediate field. In your question $G = Aut(K/F)$. Jul 3, 2019 at 22:01

No- the map is indeed injective (at least if you're dealing with fields of characteristic $$0$$).

First, some notation. For any subgroup, $$H$$, of $$\operatorname{Aut}(K\mid F)$$, denote the subset of all elements of $$K$$ fixed by every member of $$H$$, $$H^F$$ (this is just $$\operatorname{Fix}(H)$$). For any field intermediate $$K$$ and $$F$$, $$L$$, denote the subset of all elements of $$\operatorname{Aut}(K\mid F)$$ that fixes every element of $$L$$, $$L^*$$.

It is elementary to show that, for any subgroup, $$H$$, of $$\operatorname{Aut}(K\mid F)$$, $$H^F$$ is a field between $$F$$ and $$K$$, and that, for any field, $$L$$, between $$F$$ and $$K$$, $$L^*$$ is a subgroup of $$\operatorname{Aut}(K\mid F)$$.

The map from $$\operatorname{Sub}(\operatorname{Aut}(K\mid F))$$ to $$\{E\mid E=\mathrm{field}, F\leq E\leq K\}$$ under consideration here is the function that sends a subgroup, $$H$$, of $$\operatorname{Aut}(K\mid F)$$ to $$H^F$$. When $$\operatorname{char}(F)=0$$, it is injective.

In what follows, $$H$$ represents an arbitrary subgroup of $$\operatorname{Aut}(K\mid F)$$, and $$L$$, an arbitrary field between $$F$$ and $$K$$.

The first result is that: if $$H=L^*$$, then $$L\subset H^F$$.

The proof of this is pretty obvious. If $$H=L^*$$, then, by definition, every member of $$H$$ fixes all of $$L$$. $$H^F$$ is defined to be all the stuff in $$K$$ fixed by every member of $$H$$, of which, $$L$$ is clearly a part.

The second result is: if $$L=H^F$$, then $$|H|=[K:L]$$.

This one is more involved. First, we note that, as $$H$$ consists solely of $$L$$-fixing automorphisms on $$K$$, we must have $$|H|\leq [K:L]$$. Now, we use a rather complicated argument to show $$|H|\geq [K:L]$$. Combined with the first inequality, this will yield $$|H|=[K:L]$$, as desired.

Now, let the elements of $$H$$ be $$h_1,h_2,\dots,h_r$$, where $$r=|H|$$.

Since $$L\supset F$$ and $$\operatorname{char}(F)=0$$, we have $$\operatorname{char}(L)=0$$; we may then write $$K=L(a)$$ for some $$a\in K$$.

Consider the polynomial $$A(x)=\prod_{i=1}^r(x-h_i(a))$$

We note here that, since the identity automorphism appears as some $$h_i$$, $$(x-a)$$ is a factor of $$A(x)$$, i.e. $$a$$ is a root of $$A(x)$$.

For each $$i$$, define a function $$\bar h_i$$ on $$K[X]$$ so that, for any $$k(x)=k_nx^n+\dots+k_1x+k_0\in K[X]$$,
$$\bar h_i(k(x))=h_i(k_n)x^n+\dots+h_i(k_1)x+h_i(k_0)$$

It is easy to show that each $$\bar h_i$$ is an automorphism on $$K[X]$$.

Denote the group of these automorphisms on $$K[X]$$ by $$\overline{H}$$.

Now, let us apply an arbitrary $$\bar h_j$$ to $$A(x)$$: $$\bar h_j(A(x))=\bar h_j\left(\prod_{i=1}^r(x-h_i(a))\right)$$

As $$h_j$$ is an automorphism, we have $$\bar h_j(A(x))=\prod_{i=1}^r\bar h_j(x-h_i(a))$$

Using the definition of $$\bar h_j$$: $$\bar h_j(A(x))=\prod_{i=1}^r(x-h_j(h_i(a)))=\prod_{i=1}^r(x-(h_j\circ h_i)(a))$$

I have used $$\circ$$ to denote the composition of the automorphisms. Now, we note that, as $$h_j$$ is an element of $$H$$, the left coset of $$h_j$$ with $$\overline{H}$$, i.e. $$h_j\circ H$$, is simply equal to $$H$$. So the set $$\{(h_j\circ h_1)(a),(h_j\circ h_2)(a),\dots,(h_j\circ h_r)(a)\}$$ again equals $$\{h_1(a),h_2(a),\dots,h_r(a)\}$$.

Which means $$\bar h_j(A(x))=A(x)$$. Since $$j$$ was arbitrary this holds for any element of $$\overline{H}$$. But, given the definition of $$\bar h_j$$, for us to have $$\bar h_j(A(x))=A(x)$$, we must have that $$h_j$$ fixes every coefficient of $$A(x)$$. This means every member of $$H$$ fixes every coefficient of $$A(x)$$. Since we assumed $$L=H^F$$, that means $$A(x)\in L[x]$$.

Recall, however, that $$a$$ is a root of $$A(x)$$. This means the minimal polynomial of $$a$$ over $$L$$ divides $$A(x)$$. The degree of the minimal polynomial of $$a$$ over $$L$$ is $$[L(a):L]=[K:L]$$. Since this polynomial divides $$A(x)$$, we must have $$\mathrm{deg}(A(x))\geq [K:L]$$

But, by the way $$A(x)$$ was defined, $$\mathrm{deg}(A(x))=|H|$$, so we have $$|H|\geq [K:L]$$, completing the proof of the second result.

The final result is: if $$L=H^F$$, then $$L^*=H$$. If we can show this, we'll have that the map from subgroups to intermediate fields is injective. So, suppose we have $$L^*=E$$. Well, looking back at the definitions, $$E$$ is the biggest subgroup that fixes all of $$L$$. $$L$$ is the field fixed by all of $$H$$. $$H$$ fixes all of $$L$$, so we must have $$H\subset E$$.

By the $$2^{nd}$$ result, we also have $$|H|=[K:L]$$.

By the $$1^{st}$$ result, we have that $$L\subset E^F$$, and, by the $$2^{nd}$$ result, we have $$|E|=[K:E^F]$$.

Since $$L\subset E^F$$, we must have $$|E|=[K:E^F]\leq[K:L]$$, but $$|H|=[K:L]$$. So $$|E|\leq |H|$$. Combining this with the fact that $$H\subset E$$, we have $$H=E=L^*$$, as desired.

So there will always be more intermediate fields than subgroups.

(If there are any doubts, you can also explicitly show that the map fails to be surjective. Specifically, $$F$$ will never be $$H^F$$ for any subgroup, $$H$$, of $$\operatorname{Aut}(K\mid F)$$ if $$K$$ is non-Galois. Why? Suppose not. Suppose it was. What does result $$2$$ tell you then? Is this possible?)

(please comment/ edit for any corrections)