Fundamental Theorem of Galois Theory

Central Question

Why should intermediate fields and subgroups encode the same information in opposite directions?

Why This Matters

Fixed elements turn symmetries into fields, while fixing a field selects symmetries. Their correspondence is an organizing principle rather than just another theorem.

Mathematical Notes

More fixed information means fewer symmetries

Let $L/F$ be finite Galois and $G=\operatorname{Gal}(L/F)$. To an intermediate field $M$, associate the automorphisms fixing all of $M$:

To a subgroup $H\le G$, associate its fixed field:

Both maps reverse inclusion. Enlarging the field imposes more conditions on an automorphism; enlarging the subgroup imposes more conditions on a fixed element.

Why the maps really are inverse

The decisive counting result is Artin's fixed-field theorem: a finite group $H$ of field automorphisms satisfies $[L:L^H]=|H|$. Its proof uses the linear independence of distinct field embeddings; this is the substantial step behind the correspondence.

Given $M$, the extension $L/M$ is also finite Galois, so $|\operatorname{Gal}(L/M)|=[L:M]$. The fixed field of this subgroup contains $M$, and Artin's theorem gives the same degree below $L$. The tower law forces equality. Conversely, $H\subseteq\operatorname{Gal}(L/L^H)$ and both groups have order $|H|$, so they are equal.

For normality, $\sigma(L^H)=L^{\sigma H\sigma^{-1}}$. Thus the fixed field is stable under all base-field symmetries exactly when $H$ is normal. In that case restriction to $M$ is onto, has kernel $H$, and gives the quotient by the first isomorphism theorem.

The complete biquadratic example

Let $L=\mathbb Q(\sqrt2,\sqrt3)$, with $s$ flipping $\sqrt2$ and $t$ flipping $\sqrt3$.

For instance, $st$ sends $a+b\sqrt2+c\sqrt3+d\sqrt6$ to $a-b\sqrt2-c\sqrt3+d\sqrt6$. Its fixed elements have $b=c=0$. All subgroups are normal because $G$ is abelian, so all three quadratic intermediate fields are Galois over $\mathbb Q$.

What the nonabelian example adds

For $x^3-2$, $G\cong S_3$. Its order-three subgroup fixes $\mathbb Q(\omega)$; the three order-two subgroups fix $\mathbb Q(\alpha)$, $\mathbb Q(\alpha\omega)$ and $\mathbb Q(\alpha\omega^2)$. Together with the endpoints, these are all six intermediate fields.

The order-three subgroup is normal; the order-two subgroups are not. Hence the quadratic intermediate extension is Galois, but the three cubic extensions are not. Every $L/M$ is Galois; not every $M/F$ is.

Learning Prompt

Motivate the Fundamental Theorem of Galois Theory through a concrete example before stating it abstractly. Show me why subgroups of the Galois group should correspond to intermediate fields. I care much more about understanding why this correspondence exists than memorizing the theorem.

Ideas

Examples

Questions

Connections

• Field Automorphisms
• My First Genuine Galois Group
• Why the Quintic Changed Mathematics
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, Chapter 10; §12.1; §13.1. Publisher and edition details.