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I am exploring a bit, the definition of an étale algebra here's my example today... Wikiedia has that

${\displaystyle L}$ is étale if it is isomorphic to a finite product of (separable) extensions of ${\displaystyle K}$.

So I have to take a look at the definition of separable extension [1]

  • $L$ is an extension over $K$ is separable if every algebraic element $a \in L$ has a minimimal polynomial $p(x) = 0$ that is separable over $K$.
  • Theorem Any finite separable extension has the form $L = K(c)$ for some $c \in L$.
  • E.g. $\mathbb{Q}[\sqrt{2} + \sqrt{3}]$ is separable over $\mathbb{Q}$ and in fact $\mathbb{Q}[\sqrt{2}, \sqrt{3}] \simeq \mathbb{Q}[\sqrt{2} + \sqrt{3}]$
  • A non-separable extension, are those with multiple-zeros. $L = \mathbb{F}_3[x]/(x^3 - 1)$ and $K = \mathbb{F}_3$, since in that case technically $x^3 + 1 \equiv (x+1)^3$. This can only happen over finite fields.

These theorems of separable extensions look formal, encode some rather non-trivial things about the Euclidean algorithms over the ground fields (such as $\mathbb{Q}$) and over polynomial rings.


In fact, Étale algebras are also called separable algebras. I am looking for some examples: the definition has that: $$ L \otimes_K \mathbb{C} \simeq \mathbb{C}^n $$

Even these notions of separability has been intertwined with the the tensor product. Even if $K(c) \not \simeq L$: $$L \otimes_\mathbb{K} K(c) \simeq K[X]/ p(X) $$ Conrad's note gives the example that the first is a ring and the other isn't: \begin{eqnarray*} \mathbb{C}[X]/(X^2 + 1) &\simeq& \mathbb{C} \times \mathbb{C} \\ \mathbb{R}[X]/(X^2 + 1) &\simeq& \mathbb{C} \end{eqnarray*}


It remains to ask a question. Here are two more definitions of étale algebra

2 A third definition says that an étale algebra is a finite dimensional commutative algebra whose trace form $(x,y) = \mathrm{Tr}[xy]$ is non-degenerate

and

3 a finite dimensional commutative algebra over a field is étale if and only if ${\displaystyle \mathrm {Spec} \,L\to \mathrm {Spec} \,K}$ is an étale morphism.

If we let $K = \mathbb{Q}$ and $L = L_1 \times L_2$ with $L_1 = \mathbb{Q}[\sqrt[3]{2}] $ and $L_2 =\mathbb{Q}[\sqrt[3]{5}]$ do we have that $$ \mathrm{Spec} \big[\mathbb{Q}[\sqrt[3]{2}] \times \mathbb{Q}[\sqrt[7]{5}] \big] \simeq \mathrm{Spec}\big[ \mathbb{Q}[\sqrt[3]{2}] \big] \times \mathrm{Spec}\big[ \mathbb{Q}[\sqrt[3]{5}] \big] \to \mathrm{Spec}\mathbb{Q} $$ is an étale morphism? What are the points of these specs anyway? Isn't $\mathbb{Q}$ a point? What does the trace form look like?

While trivial in one definition, I would like to see a verification of either #2 or #3. It's not obvious that trace is non-degenerate, and I haven't any working idea of the definition of étale morphism.

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  • $\begingroup$ A minor comment, but the polynomial $x^3 -1$ is not irreducible over $\mathbb{F}_3$ because $1$ is a root of $x^3 -1$ mod $3$. In fact, the finite fields $\mathbb{F}_p$ are examples of perfect fields, which are fields that have the property that every extension field of $K$ is separable over $K$. Your insight about the foolishness on exponents is good, however: the field $\mathbb{F}_p(t)[x]/(x^p-t)$ is purely inseparable over $\mathbb{F}_p(t)$. $\endgroup$
    – Geoff
    Apr 6 '17 at 23:20
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A very partial answer, which ought to be superseded by other, more complete answers:

I’m guessing that your notation $R\times S$ means the set-theoretic product of $R$ and $S$ with coordinatewise addition and multiplication. When $R$ and $S$ are fields, this is a ring with only two (nonzero) proper ideals, $(0)\times S$ and $R\times(0)$, both of these being maximal.

For the associated $\mathrm{Spec}$s, you expect always to have $\mathrm{Spec}(R\times S)=\mathrm{Spec}(R)\amalg\mathrm{Spec}(S)$: that is, product of the rings corresponds to disjoint union of the spaces.

As to your question on the trace form of the product ring, I suppose that if you describe the trace form on each factor by an appropriate square matrix over the base ($\Bbb Q$ in this case), the trace form on the product ring should be a block matrix formed from the two separate matrices.

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  • $\begingroup$ it's not such a great question, so I can't expect a great answer. i did learn today maybe there's a relationship between tensor products and separable extensions. over $\mathbb{C}$ all extension are separable anyway. Maybe trouble over $p$-adic numbers. $\endgroup$
    – cactus314
    Apr 6 '17 at 22:54
  • $\begingroup$ All field extensions in characteristic zero are separable. I think that $\Bbb C[\varepsilon]/(\varepsilon^2)$ has to be considered nonseparable over $\Bbb C$. $\endgroup$
    – Lubin
    Apr 7 '17 at 2:19

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