Questions tagged [separable-algebras]

For questions about separable algebras over commutative rings, separable ring extensions, Azumaya algebras, finite etale algebras.

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Separable monads do not induce separable monoids

Let us first recall the categorical notion of monad: if we have a category $\mathcal{C}$ then a monad on it consists in an endofunctor $\mathbb{A}\colon \mathcal{C}\rightarrow \mathcal{C}$ together ...
N.B.'s user avatar
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Is a "separable" algebra over a field finite-dimensional?

Let $k$ be a field and $A$ a unital associative (possibly non-commutative) $k$-algebra, and let $A^e$ denote the enveloping algebra of $A$, namely, $A^e = A \otimes_k A^{op}$. It seems that there are ...
H. E.'s user avatar
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Explicit separability idempotent for the center of a separable algebra

Let $A$ be a $k$-algebra for some commutative ring $k$. Recall that $A$ is said to be separable over $k$ if the multiplication map $A\otimes_k A^{\operatorname{op}}\to A$ has a section as a map of $A\...
Maxime Ramzi's user avatar
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Separable algebras and separably closed local rings (a.k.a strictly Henselian local rings)

Let $A$ be a local ring. Say a monic $f\in A[x]$ is unramifiable if it admits a simple root in its universal splitting algebra (equivalently, it admits a simple root in any ring over which it splits). ...
Arrow's user avatar
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Separable nonassociative algebras

In his paper "Structure and Representation of Nonassociative Algebras", Schafer notes that an arbitrary nonassociative algebra over a field is separable "if and only if the ...
a196884's user avatar
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2 votes
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Separability of $\mathbb{C}[x,y_1,\ldots,y_r]$ over $\mathbb{C} + (h,y_1,\ldots,y_r)$

The answer to this MO question says the following: Lemma 1. Let $h \in \mathbf C[x]$ be a polynomial of degree $n \geq 2$. Then $\mathbf C+(h) \subseteq \mathbf C[x]$ is unramified if and only if $h$ ...
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Maximal orders separable over their centre

Let $\mathcal{A}$ be a central simple $K$-algebra, where $K$ is an algebraic number field. It is known that $\mathcal{A}$ is separable over $K$ (following the definition of DeMeyer and Ingraham's book)...
a196884's user avatar
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3 votes
1 answer
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Flatness of certain subrings

The following question appears, more or less, here: Let $k$ be an algebraically closed field of characteristic zero and let $S$ be a commutative $k$-algebra (I do not mind to further assume that $S$ ...
user237522's user avatar
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1 answer
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Separable non-flat simple ring extension

Let $R \subseteq S$ be two commutative $\mathbb{C}$-algebras such that: (1) $R$ and $S$ are integral domains. (2) $Q(R)=Q(S)$, namely, their fields of fractions are equal. (3) $S=R[w]$, for some $w \...
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Separability of $\mathbb{C}[x]$ over its $\mathbb{C}$-subalgebras

For commutative rings $R \subseteq S$, recall that $S$ is separable over $R$, if $S$ is a projective $S \otimes_R S$-module, via $f: S \otimes_R S \to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$. ...
user237522's user avatar
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A certain property of integral domains $A \subseteq B$ with $Q(A) \cap B= A$

I have asked the following question in MSE: Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \...
user237522's user avatar
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5 votes
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Structure theorem for etale algebras over a more general ring than a field

I call etale a finite-type flat $R$-algebra $A$ such that $\Omega_A =0$ (I hope this is the standard definition). In the case where $R=k$ is a field, any such algebra $A$ decomposes as a finite ...
display llvll's user avatar
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Classical reductive group schemes vs. unitary groups of separable algebras with involution --- reference request

Let $K$ be a field with $2\in K^\times$, let $A$ be a separable $K$-algebra (i.e. $A$ is finite-dimensional, semisimple and its center is an etale $K$-algebra), and let $\sigma:A\to A$ be a $K$-...
Uriya First's user avatar
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3 votes
1 answer
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Operation of a p'-group on a set of p-power order and fix points

The question is related to Taft's Theorem about G-invariant radical complements. Let $A$ be an associative unitary finite-dimensional $K$-Algebra posessing a separable factor Algebra by ist nilradical....
Sven Wirsing's user avatar
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0 answers
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Maximal separable subalgebras of semisimple algebras

Is anything known about maximal separable subalgebras of semisimple algebras in finite-dimension? Are those subalgebras of unique dimension or isomorphic or conjugate?
Sven Wirsing's user avatar
2 votes
0 answers
252 views

Proof of this ‘lemme connu’

In the proof of Corollary 10.12 of Exposé I of SGA 1 something like the following is asserted as a ‘known lemma’: Let $k$ be an infinite field and $B$ a finite $k$-algebra. If $B$ is not a product ...
Tomo's user avatar
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2 votes
1 answer
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Separability of a subring and of a pre-image

Let $A \subseteq B \subseteq C$ be commutative rings. A known result, which can be found in De Meyer and Ingraham's book, says that separability of $A \subseteq C$ implies separability of $B \...
user237522's user avatar
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Separability of the ring extension $A \subset A[T]/(h_1,\ldots,h_n)$

Is there a generalization of Wang's known result, Corollary 8, which says the following: $A \subset B=A[T]/(h(T))=A[w]=B$ is a separable ring extension if and only if $h'(w)$, the formal derivative of ...
user237522's user avatar
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5 votes
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What is the relationship between Frobenius extensions and Separable extensions

Let $R\to S$ be an extension of possibly non-commutative rings. I am interested in the relationship between $R\to S$ being Frobenius and it being separable. If it is a Frobenius extension, then there ...
Johannes Hahn's user avatar
13 votes
1 answer
622 views

Is a "smooth" finite-dimensional algebra separable modulo its radical?

Let $k$ be a field, and let us write the "unadorned" tensor $\otimes$ in place of $\otimes_k$. For a unital finite-dimensional $k$-algebra $A$, let $A^e = A \otimes A^{op}$ denote the enveloping ...
Manny Reyes's user avatar
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Separability and smoothness

Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable ...
user237522's user avatar
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2 votes
1 answer
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Are the fibers of this morphism geometrically regular?

Let $A\rightarrow B$ be a local morphism of complete noetherian rings making $B$ a formally smooth $A$-algebra. Does the induced morphism $\textrm{Spec}(B)\to\textrm{Spec}(A)$ have geometrically ...
Olórin's user avatar
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10 votes
1 answer
494 views

Is every separable algebra in a modular tensor category Morita equivalent to a commutative one?

Separable algebras in modular tensor categories are interesting algebraic structures, which have received significant attention because of their connection to conformal field theories. My ...
Jamie Vicary's user avatar
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Finite separable extension of fields imply the number of intermediate subfield is finite

The proof of statements either uses Galois theory or Artin primitive element theorem.I would like to know whether there is a proof without using these.The reason to avoid using Galois theory is that ...
user41650's user avatar
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7 votes
0 answers
236 views

Non Commutative Hyperspaces

Let $X$ be a compact metric space. Recall that the hyperspace $2^{X}$ is the set of all non empty compact subsets of $X$ with the Hausdorff metric. Assume that $\mathcal{C}$ is the category of all ...
Ali Taghavi's user avatar
4 votes
2 answers
536 views

An R-algebra A is R-separable if and only if all derivations are inner.

Hello everybody. I'm readying about derivations. It is very very known fact that all derivations $\delta: A\rightarrow M$ (A R-algebra, M A-module) are inner when the algebra is R-separable. ...
muestrass's user avatar
2 votes
1 answer
386 views

On the Separability of Certain Extensions of Fields

I asked this question a couple of weeks ago. The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $...
David's user avatar
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7 votes
1 answer
399 views

Hochschild H^1 (R,M) = 0 vs. H_1 (R,M) = 0 where R is a ring and M is an (R,R)-bimodule

Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative). Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the ...
darij grinberg's user avatar