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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 …

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 …

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 …

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 …

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 …

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 …

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. …

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 …

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 …

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 …

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 …

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 …