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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
5
votes
Accepted
0 dimensional Dedekind domain?
Historically Emmy Noether's paper introducing the concept "Dedekind domain" certainly included fields (see e.g. Kleiner's book on the history of abstract algebra, which gives axioms). She was characte …
4
votes
Accepted
Decomposition of finite algebras over finite fields
The general form of the answer is easy to anticipate: a finite commutative ring is artinian. It will be a product of finite local rings, of characteristic that is a prime number. So your question is r …
4
votes
Is an elementary symmetric polynomial an irreducible element in the polynomial ring?
Doesn't this follow quite quickly by setting one variable equal to 0?
Edit: I was thinking this way. Factors of homogeneous polynomials are homogeneous. Setting the final variable $x_n$ to 0 therefo …
2
votes
Who named it the Snake Lemma?
If you Google for "diagramme du serpent" it becomes plausible that it was a diagram in Cartan-Eilenberg first of all, before a lemma. Interesting example of how Bourbaki became the standard grad stude …
2
votes
Generators of a maximal ideal of $k[X_1,\cdots,X_n]$
The dimension is exactly n: you are counting linear polynomials modulo terms of higher order.
The geometrical interpretation is that intersecting n - 1 hypersurfaces with a common point must give an …
2
votes
Is there a field which is the union of finitely many proper subfields?
The reason it fails in the case of finite fields is the primitive element theorem (over the prime subfield, even). This is not quite an argument in the general case because of possible inseparability. …
1
vote
Is $x^p-x+1$ always irreducible in $\mathbb F_p[x]$?
There is kind of an easy proof given that Berlekamp's algorithm works?
In the notation of https://en.wikipedia.org/wiki/Berlekamp%27s_algorithm, the point is that the space of polynomials g congruen …
1
vote
Subrings of rational functions invariant under change of sign
So at the level of function fields, you have a group of order 2 acting on L, the field of fractions of R, and there is a subfield K fixed by it. The extension L/K will be of degree 2, so quadratic. In …
1
vote
Elementary Luroth theorem proof?
I think this is something Gauss could have proved, and the point is to come up with his sort of proof. I'm not seeing that as too hard. To show a polynomial of degree at least 1 is transcendental over …
1
vote
cyclic polynomials and their solutions
So you have a cyclic group acting on complex affine space, and the situation at the level of function fields is clear enough: by basic Galois theory the extension of rational functions is a cyclic ext …
0
votes
Is there much difference between Kronecker's and Dedekind's methods in algebraic number theo...
Urgh - Hilbert stole Kronecker's key ideas on "module theory" to use in invariant theory, while excluding them from his Zahlbericht, and trying to wipe out the "constructive" point of view? Dedekind w …
0
votes
Accepted
How to design or create or generate a bijective ring map?
In generality (this is tagged "commutative algebra", so let's talk commutative rings) I wonder if there is more than taking generators of each side and writing the images as polynomials in the generat …