Étale Algebras

Suppose that a $\mathbb{K}$-algebra $A$ and a field extension $\mathbb{L}/\mathbb{K}$ are given. Then the collection of $\mathbb{K}$-linear maps between $\mathbb{K}$-vector spaces, $\Hom_\mathbb{K}(A, \mathbb{L})$, is a $\mathbb{K}$-vector space, and via the isomorphism

\[\Hom_\mathbb{K}(A,\mathbb{L})\cong\Hom_\mathbb{K}(A, \mathbb{K}\otimes_\mathbb{K}\mathbb{L})\cong\Hom_\mathbb{K}(A, \mathbb{K})\otimes_\mathbb{K}\mathbb{L}=A^\ast\otimes_\mathbb{K}\mathbb{L}\]

it can also be regarded as an $\mathbb{L}$-vector space. ([Multilinear Algebra] §Hom and the Tensor Product, ⁋Proposition 3)

Slightly changing the order of the above process, for $A_{(\mathbb{L})}=\mathbb{L}\otimes_\mathbb{K}A$, let us consider the dual (as an $\mathbb{L}$-vector space) $(A_{(\mathbb{L})})^\ast$. Then, since the Hom-tensor adjunction

\[(A_{(\mathbb{L})})^\ast=\Hom_\mathbb{L}(\mathbb{L}\otimes_\mathbb{K}A, \mathbb{L})\cong\Hom_\mathbb{K}(A, \Hom_\mathbb{L}(\mathbb{L}, \mathbb{L}))\cong\Hom_\mathbb{K}(A, \mathbb{L})\]

exists, from this we obtain an isomorphism between $\mathbb{L}$-vector spaces:

\[(A_{(\mathbb{L})})^\ast=\Hom_\mathbb{L}(A_{(\mathbb{L})}, \mathbb{L})\cong\Hom_\mathbb{K}(A,\mathbb{L})\cong A^\ast\otimes_\mathbb{K}\mathbb{L}\]

In particular, if $A$ is finite-dimensional as a $\mathbb{K}$-vector space, then $A^\ast$ is a $\mathbb{K}$-vector space of the same dimension, and therefore $A_{(\mathbb{L})}$ and $A_{(\mathbb{L})}^\ast$ are also $\mathbb{L}$-vector spaces of the same dimension. In particular, we obtain the following equality:

\[[A_{(\mathbb{L})}:\mathbb{L}]=\dim_\mathbb{L}A_{(\mathbb{L})}=\dim_\mathbb{L} (A_{(\mathbb{L})})^\ast=\dim_\mathbb{K}A=[A:\mathbb{K}]\]

From this, we obtain the following key ideas of this post.

The extension degree $[A:\mathbb{K}]$ behaves well under base change.
To compute the extension degree $[A:\mathbb{K}]$, it suffices to take any extension $\mathbb{L}/\mathbb{K}$ and compute the dimension of $\Hom_\mathbb{K}(A,\mathbb{L})$.

Then the following theorem is related to the second idea, and helps compute the dimension of $\Hom_\mathbb{K}(A,\mathbb{L})$, which is the object of our interest.

Theorem 1 Let an extension $\mathbb{L}/\mathbb{K}$ be given and fix a $\mathbb{K}$-algebra $A$. Then $\Hom_{\Alg{\mathbb{K}}}(A,\mathbb{L})$ is a free subset of the $\mathbb{L}$-vector space $\Hom_\mathbb{K}(A, \mathbb{L})$.

Proof

That is, it suffices to show that any finite subset of $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$, $\{u_1,\ldots, u_n\}$, is necessarily linearly independent. We use induction on $n$. The case $n=0$ is trivial, so assume $n \geq 1$.

For elements of $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$, $u_1,\ldots, u_n$, suppose that elements of $\mathbb{L}$, $\alpha_1,\ldots,\alpha_n$, satisfy the following equation:

\[\sum_{i=1}^{n} \alpha_i u_i = 0\]

Then for $x,y\in A$, from the computation

\[\sum_{i=1}^{n-1} \alpha_i(u_i(x) - u_n(x)) u_i(y) = \sum_{i=1}^{n} \alpha_i u_i(xy) - u_n(x) \sum_{i=1}^{n} \alpha_i u_i(y) = 0\]

we see that

\[\sum_{i=1}^{n-1} \alpha_i(u_i(x) - u_n(x)) u_i = 0\]

Now by the inductive hypothesis, we see that the coefficients $\alpha_i(u_i(x)-u_n(x))$ in the above equation must all be $0$. However, $u_1,\ldots, u_n$ are all distinct elements, and therefore, regardless of the value of $x$, for this to be identically $0$, we must have $\alpha_i=0$ for all $i=1,\ldots, n-1$. Substituting this back into the equation of the assumption, we see that $\alpha_n$ must also be $0$.

This theorem itself is not so surprising: for instance, given $\Hom_\mathbb{K}(\mathbb{K},\mathbb{K})$ and the subset $\Hom_{\Alg{\mathbb{K}}}(\mathbb{K},\mathbb{K})$, for any $\lambda\in \mathbb{K}$ the following function

\[x\mapsto \lambda x\]

is always a $\mathbb{K}$-linear map, but for it to be a $\mathbb{K}$-algebra homomorphism, the equation $\lambda(xy)=\lambda(x)\lambda(y)$ must hold, so we must have $\lambda^2=\lambda$, and therefore $\Hom_{\Alg{\mathbb{K}}}(\mathbb{K},\mathbb{K})$ consists of only the single element $\id$.

In any case, from the above theorem we obtain the following inequality:

\[\lvert\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})\rvert\leq \dim_\mathbb{L}\Hom_\mathbb{K}(A,\mathbb{L})=[A:\mathbb{K}]\]

In particular, we obtain the following two corollaries.

Corollary 2 Fix a monoid $\Gamma$ and a field $\mathbb{L}$, and let $X$ be the set of homomorphisms from $\Gamma$ to $\mathbb{L}^\times$. Then $X$ is a free subset of the space of functions from $\Gamma$ to $\mathbb{L}$, the $\mathbb{L}$-vector space $\mathbb{L}^\Gamma$.

Proof

Consider the monoid algebra $A=\mathbb{L}\Gamma$, and consider the canonical basis $(e_\gamma)_{\gamma\in\Gamma}$.

Then through a generalization of [Algebraic Structures] §Algebras, ⁋Proposition 6, there is a bijection between $X$ and $\Hom_\mathbb{L}(A,\mathbb{L})$, so the claim follows immediately from Theorem 1.

Corollary 3 (Dedekind) For two extensions $\mathbb{L}/\mathbb{K}$ and $\mathbb{M}/\mathbb{K}$, the set of morphisms from $\mathbb{M}$ to $\mathbb{L}$ is free as an $\mathbb{L}$-vector space. In particular, if $\mathbb{M}/\mathbb{K}$ is a finite degree extension, the number of elements of this set is at most $[\mathbb{M}:\mathbb{K}]$.

Proof

It suffices to set $A=\mathbb{M}$ and use Theorem 1. The second assertion is obvious because $\Hom_\mathbb{K}(\mathbb{M},\mathbb{L})$ is an $\mathbb{L}$-vector space of dimension $[\mathbb{M}:\mathbb{K}]$.

Furthermore, if $\mathbb{K}$ is an infinite set, these are also algebraically independent.

Theorem 4 Let an infinite field $\mathbb{K}$ and an extension $\mathbb{L}/\mathbb{K}$ be given, and fix a $\mathbb{K}$-algebra $A$. If for distinct $\mathbb{K}$-algebra homomorphisms $u_1,\ldots, u_n:A \rightarrow \mathbb{L}$, a polynomial $f\in \mathbb{L}[\x_1,\ldots, \x_n]$ satisfies $f(u_1,\ldots, u_n)=0$ identically, then $f=0$.

Proof

Let the subset of elements of the form $(u_1(x), \dots, u_n(x))$ in $\mathbb{L}^n$ be $B$. Then from Theorem 1, we know that to satisfy

\[\sum_{i=1}^n\alpha_i u_i(x)=0\]

for all $x\in A$, no non-$0$ $(\alpha_1,\ldots, \alpha_n)$ exists. In other words, if we define the bilinear pairing $B\times \mathbb{L}^n \rightarrow \mathbb{L}$ by

\[\bigl((u_1(x),\cdots, u_n(x)), (\alpha_1,\ldots, \alpha_n)\bigr) \mapsto \sum_{i=1}^n \alpha_iu_i(x)\]

then the induced map $\mathbb{L}^n\rightarrow B^\ast$ is injective, and from this we know that $B$ must generate $\mathbb{L}^n$. Therefore, so that the $n\times n$ matrix $(u_i(a_j))$ is invertible, there exist elements $a_j\in A$.

Now define the polynomial $g \in \mathbb{L}[\y_1,\ldots, \y_n]$ by the following formula:

\[g(\y_1, \ldots, \y_n) = f\left( \sum_{j=1}^n u_1(a_j)\y_j, \ldots, \sum_{j=1}^n u_n(a_j)\y_j \right)\]

Substituting arbitrary elements $y_i\in \mathbb{K}$ here and setting $x=\sum_{i=1}^n a_iy_i$, we have

\[g(y_1, \dots, y_n) = f(u_1(x), \dots, u_n(x)) = 0\]

and from the hypothesis on $f$, we know that $g(y_1,\ldots, y_n)=0$. Now since $\mathbb{K}$ is an infinite set, $g$ must be identically $0$, and letting the inverse matrix of $(u_i(a_j))$ be $(v_{ij})$, we have

\[f(\x_1,\ldots, \x_n)=g\left(\sum_{j=1}^n v_{1j}\x_j, \dots, \sum_{j=1}^n v_{nj}\x_j \right)\]

and therefore $f$ is also identically $0$.

Étale Algebras

Definition 5 A $\mathbb{K}$-algebra $A$ is diagonalizable if there exists a suitable $n\geq 0$ such that there is a $\mathbb{K}$-algebra isomorphism $A\cong \mathbb{K}^n$. If for some $n\geq 0$, an $\mathbb{L}$-algebra isomorphism $A_{(\mathbb{L})}\cong \mathbb{L}^n$ exists for an extension $\mathbb{L}/\mathbb{K}$, we say that this extension diagonalizes $A$. If $A$ is diagonalized by a suitable extension $\mathbb{L}/\mathbb{K}$, we call $A$ an étale algebra.

A diagonalizable algebra is useful in that its extension degree $[A:\mathbb{K}]$ is well defined. Using the first key idea explained at the beginning of this post, we can say that an étale algebra is just as useful.

First, we begin with the following characterization of diagonalizable algebras.

Proposition 6 For a finite degree $n$ $\mathbb{K}$-algebra $A$, the following are all equivalent.

$A$ is diagonalizable.
For $A$, there exists a suitable basis $(e_1,\ldots, e_n)$ such that $e_i^2=e_i$ and, for $i\neq j$, $e_ie_j=0$.
From $A$ to $\mathbb{K}$, the $\mathbb{K}$-algebra homomorphisms generate $A^\ast$.
Every $A$-module decomposes as a direct sum of submodules that are one-dimensional over $\mathbb{K}$.

Proof

The equivalence of the first two conditions follows immediately from the multiplication structure defined on $\mathbb{K}^n$. On the other hand, the projections from $\mathbb{K}^n$ to $\mathbb{K}$, $n$ in number, are $\mathbb{K}$-algebra homomorphisms, so these conditions imply the third condition.

Conversely, assume that the third condition holds, and among the elements generating $A^\ast$ that are $\mathbb{K}$-algebra homomorphisms, let us choose a basis $u_1,\ldots, u_n$. Then one can show that $x\mapsto (u_i(x))$ is an isomorphism from $A$ to $\mathbb{K}^n$ of $\mathbb{K}$-algebras, and therefore the first three conditions are all equivalent.

Now we must show the equivalence of the fourth condition. To this end, assume that the second condition holds, and pick an arbitrary $A$-module $M$. Then, considering the endomorphism defined by $x\mapsto e_ix$ on $M$ as an $A$-endomorphism, $M$ is the direct sum of the $e_iM$, from which the fourth condition follows. Conversely, assuming that the fourth condition holds, we can in particular write $A$ itself as an internal direct sum of one-dimensional $\mathbb{K}$-vector spaces, and one can verify that the basis in this case satisfies the condition of the second equivalence.

In particular, the fourth condition provides some justification for calling $A$ diagonalizable. The following corollary tells us when the set $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ of Theorem 1 becomes a basis.

Corollary 7 For a field extension $\mathbb{L}/\mathbb{K}$ and a finite degree $\mathbb{K}$-algebra $A$, consider the collection of $\mathbb{K}$-algebra homomorphisms $A \rightarrow \mathbb{L}$, $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$. Then equality in the inequality

\[\lvert \Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})\rvert \leq [A:\mathbb{K}]\]

holds if and only if $A$ is diagonalized by $\mathbb{L}$, and in this case $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ forms a basis of $\Hom_\mathbb{K}(A,\mathbb{L})$.

Proof

We already know that $\dim_\mathbb{L}\Hom_\mathbb{K}(A,\mathbb{L})=\dim_\mathbb{K}A$, and by Theorem 1 we know that $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ is a free subset of $\Hom_\mathbb{K}(A,\mathbb{L})$. Therefore, the inequality in the claim is obvious, and equality holds only when $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ is a basis of $\Hom_\mathbb{K}(A,\mathbb{L})$.

Meanwhile, the $\mathbb{L}$-vector space isomorphism $\Hom_\mathbb{K}(A,\mathbb{L}) \rightarrow (A_{(\mathbb{L})})^\ast$ sends the subset $\Hom_\Alg{\mathbb{K}}(A, \mathbb{L})$ to the set of algebra homomorphisms $A_{(\mathbb{L})} \rightarrow \mathbb{L}$, $\Hom_\Alg{\mathbb{L}}(A_{(\mathbb{L})}, \mathbb{L})$. Now, by the third equivalent condition of Proposition 6, the fact that this set $\Hom_\Alg{\mathbb{L}}(A_{(\mathbb{L})}, \mathbb{L})$ generates $(A_{(\mathbb{L})})^\ast$ is equivalent to $A$ being diagonalized by $\mathbb{L}$, which yields the desired result.

That is, if $A$ is an étale algebra, then the extension degree of $A$ over $\mathbb{K}$ equals the number of elements of $\Hom_{\Alg{\mathbb{K}}} (A, \mathbb{L})$, and carefully examining this proof shows that what we are actually calculating through this is, for an extension that diagonalizes $A$, namely $\mathbb{L}/\mathbb{K}$, the degree $[A_{(\mathbb{L})}:\mathbb{L}]$. From this perspective, the following holds.

Proposition 8 For a $\mathbb{K}$-algebra $A$, the following are all equivalent.

$A$ is étale.
There exists a finite degree extension that diagonalizes $A$.
$\overline{\mathbb{K}}/\mathbb{K}$ diagonalizes $A$.

Proof

First assume the first condition, let $n=[A:\mathbb{K}]$, and suppose that $\mathbb{L}/\mathbb{K}$ diagonalizes $A$. Then by Corollary 7, the size of $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ is $n$. On the other hand, for any $u\in \Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$, it is obvious that $[u(A):\mathbb{K}]\leq n$, and therefore the images of the elements of $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ generate in $\mathbb{L}$ a subextension $\mathbb{L}'$ that is of finite degree over $\mathbb{K}$. Now, since the number of distinct homomorphisms from $A$ to $\mathbb{L}'$ is $n$, again by Corollary 7, $\mathbb{L}'$ diagonalizes $A$.

Now, that the second condition implies the third condition follows immediately from the fact that any finite degree extension $\mathbb{L}/\mathbb{K}$ can be viewed as a subextension of $\overline{\mathbb{K}}$, and that the third condition implies the first condition is just by definition.

The finiteness in the next proposition justifies the name étale.

Proposition 9 For an étale $\mathbb{K}$-algebra $A$, $A$ has only finitely many subalgebras and ideals. Moreover, any extension that diagonalizes $A$ also diagonalizes the subalgebras and quotient algebras of $A$, and therefore every subalgebra and quotient algebra of $A$ is étale.

Proof

It suffices to show that $\mathbb{K}^n$ has only finitely many subalgebras and ideals, and that its subalgebras and quotient algebras are diagonalizable.

Let the canonical basis of $\mathbb{K}^n$ be $(e_1, \dots, e_n)$, and for a subalgebra of $\mathbb{K}^n$, $A$, let the restrictions of the projection maps $\mathbb{K}^n \rightarrow \mathbb{K}$ to $A$ be $v_1,\ldots, v_n$. Then the intersection of their kernels is $0$, so the $v_i$ generate (as a $\mathbb{K}$-vector space) $A^\ast$, and therefore $A$ is diagonalizable.

Therefore, every subalgebra of $A$ is also diagonalizable, and so whenever a subalgebra of $A$ is given, we must be able to find a basis satisfying the second condition of Proposition 6. However, the idempotents of $\mathbb{K}^n$ are precisely only those that, for a subset of $\{1,\ldots, n\}$, $I$, appear in the form $e_I=\sum_{i\in I} e_i$, and these satisfy $e_Ie_J=e_{I\cap J}$. That is, the families of idempotents satisfying the second condition are at most as many as the number of partitions of $\{1,\ldots, n\}$, and therefore $\mathbb{K}^n$ has only finitely many subalgebras.

Similarly, letting $\mathfrak{a}_I$ be the subspace having $(e_i)_{i\in I}$ as a basis of $\mathbb{K}^n$, we can show the finiteness of ideals, and the diagonalizability of $\mathbb{K}^n/\mathfrak{a}_I$ can also be shown by Proposition 6.

Separable Degree

Now, before looking further into the properties of étale algebras, let us introduce a useful concept. In the discussions so far, the set $\Hom_{\Alg{\mathbb{K}}}(A, \mathbb{L})$ that first appeared in Theorem 1 has played an important role. Now fix a commutative $\mathbb{K}$-algebra $A$ of finite degree, and whenever an extension $\mathbb{L}/\mathbb{K}$ is given, define the natural number $h(\mathbb{L})=\lvert \Hom_{\Alg{\mathbb{K}}}(A,\mathbb{L})\rvert$. Then we know that the inequality

\[h(\mathbb{L})\leq [A:\mathbb{K}]=n\]

always holds. Moreover, considering the third condition of Proposition 8, if there exists an $\mathbb{L}$ that makes the above inequality an equality, we know that $h(\overline{\mathbb{K}})$ must also make it an equality. Motivated by this, we make the following definition.

Definition 10 We define the natural number $h(\overline{\mathbb{K}})$ to be the separable degree of $A$ and write it as $[A:\mathbb{K}]_s$.

Of course, for this to be well-defined, this value must not depend on the choice of $\overline{\mathbb{K}}$. This is a consequence of the following lemma.

Lemma 11 Fix any algebraic closure $\Omega$ of $\mathbb{K}$. Then for any extension $\mathbb{L}/\mathbb{K}$, we have $h(\mathbb{L})\leq h(\Omega)$.

Proof

Let $\mathbb{L}'$ be the algebraic closure of $\mathbb{K}$ in $\mathbb{L}$. Then for any homomorphism $u:A \rightarrow \mathbb{L}$,

\[[u(A):\mathbb{K}]\leq n\]

holds. That is, $u(A)$ is an algebraic extension and thus is contained in $\mathbb{L}'$. From this we see that $h(\mathbb{L}')=h(\mathbb{L})$ must hold.

On the other hand, by §Algebraic Closure, ⁋Proposition 11, $\mathbb{L}'$ is isomorphic to a suitable subextension of $\Omega$, and therefore

\[h(\mathbb{L})=h(\mathbb{L}')\leq h(\Omega)\]

holds.

From this, if $\mathbb{L}$ is algebraically closed, we can interchange the roles of $\mathbb{L}$ and $\Omega$, so equality must hold; therefore Definition 10 is well-defined. The following are basic properties concerning this.

Proposition 12 The following hold.

For any finite degree $\mathbb{K}$-algebras $A,B$, we have $[A\otimes_\mathbb{K}B:\mathbb{K}]_s=[A:\mathbb{K}]_s[B:\mathbb{K}]_s$.
For any extension $\mathbb{K}'/\mathbb{K}$ and any finite degree $\mathbb{K}$-algebra $A$, we have $[A_{(\mathbb{K}')}:\mathbb{K}']_s=[A:\mathbb{K}]_s$.
For any finite degree extension $\mathbb{K}'/\mathbb{K}$ and any finite degree $\mathbb{K}'$-algebra $A'$, we have $[A':\mathbb{K}]_s=[A':\mathbb{K}']_s[\mathbb{K}':\mathbb{K}]_s$.

Proof

Fix an algebraic closure $\mathbb{L}$ of $\mathbb{K}$ and consider the three sets
\[\Hom_\Alg{\mathbb{K}}(A, \mathbb{L}),\qquad \Hom_\Alg{\mathbb{K}}(B, \mathbb{L}),\qquad \Hom_\Alg{\mathbb{K}}(A\otimes_\mathbb{K}B, \mathbb{L})\]
Then the given claim is clear from the fact that the map
\[\Hom_\Alg{\mathbb{K}}(A,\mathbb{L})\times \Hom_\Alg{\mathbb{K}}(B, \mathbb{L}) \rightarrow \Hom_\Alg{\mathbb{K}}(A\otimes_\mathbb{K}B,\mathbb{L}); (u,v)\mapsto u\otimes v\]
is a bijection.
Let $\mathbb{L}$ be an algebraic closure of $\mathbb{K}'$. Then the morphism
\[\bar{u}(x) = u(1 \otimes x) \quad (x \in A)\]
defines a bijection between the set of $\mathbb{K}'$-algebra homomorphisms $u$ and the set of $\mathbb{K}$-algebra homomorphisms $\bar{u}$. Therefore the equality holds.
Similarly, let $\mathbb{L}$ be an algebraic closure of $\mathbb{K}'$. Consider the two sets $\Hom_{\Alg{\mathbb{K}}}(\mathbb{K}', \mathbb{L})$ and $\Hom_\Alg{\mathbb{K}}(A', \mathbb{L})$. Now for each $u\in\Hom_{\Alg{\mathbb{K}}}(\mathbb{K}', \mathbb{L})$, consider the set
\[\Hom_\Alg{\mathbb{K}}(A', \mathbb{L})_u=\left\{v\in \Hom_{\Alg{\mathbb{K}}}(A', \mathbb{L})\mid\text{$v(x.1)=u(x)$ for all $x\in \mathbb{K}'$}\right\}\]
These induce the partition
\[\Hom_\Alg{\mathbb{K}}(A', \mathbb{L})=\bigsqcup_{u\in \Hom_\Alg{\mathbb{K}}(\mathbb{K}', \mathbb{L})} \Hom_\Alg{\mathbb{K}}(A', \mathbb{L})_u\]
On the other hand, for each fixed $u$, since $\mathbb{L}$ is an algebraic closure of $\mathbb{K}'$, the above set $\Hom_\Alg{\mathbb{K}}(A', \mathbb{L})_u$ has size $[A':\mathbb{K}']_s$ by definition. From this we obtain the desired equality.

In this language, Corollary 7 translates as follows.

Proposition 13 For a finite degree commutative $\mathbb{K}$-algebra $A$, we have $[A:\mathbb{K}]_s\leq [A:\mathbb{K}]$, and equality holds when $A$ is an étale algebra.

In particular, combining this with Proposition 12, we obtain the following corollary.

Corollary 14 The following hold.

For any two finite-degree commutative $\mathbb{K}$-algebras $A,B$, $A\otimes_\mathbb{K}B$ is étale if and only if each of them is étale.
For any extension $\mathbb{K}'/\mathbb{K}$, a finite-degree $\mathbb{K}$-algebra $A$ is étale if and only if $A_{(\mathbb{K}')}$ is étale.
For any extension $\mathbb{K}'/\mathbb{K}$ and any finite-degree $\mathbb{K}'$-algebra $A'$, $A'$ is étale over $\mathbb{K}$ if and only if $A'$ is étale over $\mathbb{K}'$ and $\mathbb{K}'$ is étale over $\mathbb{K}$.

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Étale Algebras

Étale Algebras

Separable Degree

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