Galois theory diamond diagram

hbghlyj

Let $E/F$ be finite Galois with $L_1,L_2\subseteq E$. Put
$$H_i=\mathrm{Gal}(E/L_i)\subseteq G=\mathrm{Gal}(E/F)$$
Then the Galois correspondence gives the anti-isomorphism of lattices
$$
E^{H_1\cap H_2}=L_1L_2,\qquad E^{\langle H_1,H_2\rangle}=L_1\cap L_2.
$$
So “meet of subgroups = compositum of fields”, and “generated subgroup = intersection of fields”.\[
\xymatrix{
&E\ar@{-}[d]\\
& L_1L_2 \ar@{-}[dl] \ar@{-}[dr] & \\
L_1 \ar@{-}[dr] & & L_2 \ar@{-}[dl] \\
& L_1 \cap L_2\ar@{-}[d] &\\
&F
}
\]Degree formula:
$$[L_1 L_2 : F] \cdot [L_1 \cap L_2 : F]=[L_1 : F] \cdot [L_2 : F].$$
This is sometimes called the lattice formula for Galois extensions.
(It’s the field-theoretic analog of the product formula for the orders of subgroups in a finite group.)

---

The diamond degree inequality (no Galois assumption)
For finite extensions $L_1/F$ and $L_2/F$,
$$
[L_1L_2:F]\,[L_1\cap L_2:F]\ \le\ [L_1:F]\,[L_2:F] \tag{*}
$$
A quick proof: use towers and the basic inequality
$$
[L_1L_2:L_1]\ \le\ [L_2:L_1\cap L_2].
$$
Multiplying by $[L_1:F]$ and rewriting $[L_2:F]=[L_2:L_1\cap L_2][L_1\cap L_2:F]$ yields (*).

(Why the basic inequality? For instance, pick a basis of $L_2$ over $L_1\cap L_2$; it spans $L_1L_2$ over $L_1$, so $[L_1L_2:L_1]$ can’t exceed its size.)

hbghlyj

For $K=L_1\cap L_2$ and a $K$-basis $\{b_1,\dots,b_n\}$ of $L_2$, we showed $\{1\otimes b_i\}$ is an $L_1$-basis of $L_1\otimes_K L_2$.
The multiplication map$$\mu:\;L_1\otimes_K L_2\to L_1L_2,\qquad x\otimes y\mapsto xy$$
is $L_1$-linear (where $L_1L_2$ is viewed as an $L_1$-vector space).
Hence the images $\{\mu(1\otimes b_i)=b_i\}$ span $L_1L_2$ over $L_1$. So
$$[L_1L_2:L_1]\le n=[L_2:K].$$
Finally multiply both sides by $[L_1:F]$ and use $[L_2:F]=[L_2:K][K:F]$ to get the diamond degree inequality
$$
[L_1L_2:F]\,[K:F]\le [L_1:F]\,[L_2:F].
$$equality is equivalent to injectivity of $μ$ (linear disjointness)

hbghlyj

Take a finite Galois extension $E/F$ that contains $L_1$ and $L_2$ (we can pass to a common finite Galois closure). Put
$$
G=\mathrm{Gal}(E/F),\qquad H_i=\mathrm{Gal}(E/L_i)\ (i=1,2).
$$
Then by the Galois correspondence
$$
L_1L_2 = E^{H_1\cap H_2},\qquad K=L_1\cap L_2 = E^{\langle H_1,H_2\rangle}.
$$
Now if $L_1/F$ is Galois, then $H_1$ is a normal subgroup of $G$. For normal subgroups we have
$$
\langle H_1,H_2\rangle = H_1 H_2 \quad(\text{product subgroup}),
$$
and the group-index formula gives
$$
[G:H_1\cap H_2]\,[G:H_1 H_2] = [G:H_1]\,[G:H_2].
$$
Translating these indices back into field-degrees (index = degree of fixed field over $F$) yields
\begin{align*}
[L_1L_2:F]\,[K:F] &= [G:H_1\cap H_2]\,[G:\langle H_1,H_2\rangle] \\
&= [G:H_1\cap H_2]\,[G:H_1H_2] \\
&= [G:H_1]\,[G:H_2] \\
&= [L_1:F]\,[L_2:F].
\end{align*}
i.e. the diamond equality. Thus if one of the two extensions (here $L_1$) is Galois over $F$, equality holds — equivalently $L_1$ and $L_2$ are linearly disjoint over $K$.

hbghlyj

Let $F=\mathbb Q$. Let $\alpha=\sqrt[3]{2}$ and let $\omega$ be a primitive cube root of unity. Put
$$
L_1=\mathbb Q(\alpha),\qquad L_2=\mathbb Q(\omega\alpha).
$$
Both $L_1$ and $L_2$ are degree $3$ over $\mathbb Q$: $[L_1:\mathbb Q]=[L_2:\mathbb Q]=3$ because each is generated by a root of $x^3-2$, which is irreducible over $\mathbb Q$.

Their common Galois closure is the splitting field of $x^3-2$,
$$
E=\mathbb Q(\alpha,\omega)=\mathbb Q(\alpha,\omega\alpha),
$$
and $[E:\mathbb Q]=6$ (well-known — the Galois group is $S_3$). Now check the four parts of the diamond:

• The intersection $K:=L_1\cap L_2$ is $\mathbb Q$. (No nontrivial subfield is shared — the two cubic subfields are different embeddings of the same irreducible polynomial into $E$.)
• The compositum $L_1L_2$ is the whole splitting field $E$, so $[L_1L_2:\mathbb Q]=6$.

Now compute the inequality:
$$
[L_1L_2:\mathbb Q]\,[K:\mathbb Q]=6\cdot 1 = 6
\qquad\text{while}\qquad
[L_1:\mathbb Q]\,[L_2:\mathbb Q]=3\cdot 3 = 9.
$$
So we have a strict inequality
$$
6<9,
$$
i.e. $[L_1L_2:F][L_1\cap L_2:F] < [L_1:F][L_2:F]$.
That shows the tensor map $\mu:L_1\otimes_{\mathbb Q}L_2\to L_1L_2$ is not injective — a kernel element is:
$$\theta=2(1\otimes 1)+\alpha\otimes(\omega\alpha)^2+\alpha^2\otimes(\omega\alpha)$$

• Check $\mu(\theta)=0$.
  Apply the multiplication map $\mu(a\otimes b)=ab$:
  $$
  \mu(\theta)=2 + \alpha\cdot(\omega\alpha)^2 + \alpha^2\cdot(\omega\alpha)
  =2 + \omega^2\alpha^3 + \omega\alpha^3.
  $$
  Since $\alpha^3=2$ and $\omega^2+\omega=-1$,
  $$
  \mu(\theta)=2 + 2(\omega^2+\omega)=2 + 2(-1)=0.
  $$
• Check $\theta\neq0$.
  Write $\theta$ in the $\mathbb Q$-basis$$
  \mathcal B=\{\alpha^i\otimes(\omega\alpha)^j:\;i,j=0,1,2\}
  $$Its coordinates are
  $$
  c_{00}=2,\qquad c_{12}=1,\qquad c_{21}=1,
  $$
  and all other $c_{ij}=0$. Because $\mathcal B$ is a $\mathbb Q$-basis of $L_1\otimes_{\mathbb Q}L_2$, the representation of an element in these basis vectors is unique. The coordinates are not all zero, hence $\theta\neq0$ in $L_1\otimes_{\mathbb Q}L_2$.

So $\theta$ is a nonzero element of $\ker\mu$; this explicitly shows $\mu$ is not injective and $L_1,L_2$ are not linearly disjoint over $F$.