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2d
comment Can we find the GCD of two polynomials in $\mathbb Q[x]$ by representing the coefficients as vectors?
In Algebra the polynomial ring over a commutative ring $R$ is defined as the set of sequences in $R$, that eventually become constant equal to $0$. Therefore all algebraic operations must be expressible in terms of vectors, of varying length of course, as Jyrki already pointed out in the case of the Euclidean algorithm.
2d
answered Primitive element and field extension
Jan
22
answered Two discrete lines always intersect at a point
Jan
22
comment Two discrete lines always intersect at a point
What does "extension of the affine space" mean? Projective space? What does the adjective "discrete" mean? Different?
Jan
22
answered Intersection points of curves
Jan
19
comment Problem involving a tower of fields with an algebraic and a normal extension
In general $x\in K$; of course $x\in F$ is allowed, but then $\sigma (x)=x$ and there is nothing to prove.
Jan
15
answered Problem involving a tower of fields with an algebraic and a normal extension
Dec
18
answered Is it possible to get a neighborhood with only finitely many points in it, in an infinite set?
Dec
9
awarded  Caucus
Dec
4
comment Showing a map is a bounded linear operator.
Steps for solving this exercise could be: 1. show that $A$ is linear, 2. compute the operator norm, 3. if the norm is finite conclude that $A$ is continous.
Dec
1
answered Which general methods of field construction do we know?
Nov
23
comment On Galois closure
Yes, just work over the fixed field $M^\sigma$ of $\sigma$ in the normal closure $M|F$ of $K|F$. The extension $M|M^\sigma$ is Galois, and if you can prove that it is cyclic, then $\mathrm{Aut}_F(K)$ is cyclic, because the purely inseparable part $M^\sigma|F$ does not change the automoorphism group.
Nov
23
comment On Galois closure
According to your definition cyclic extensions are allowed to be inseparable. Thus you should not work with the Galois closure but with the normal closure. The normal closure of a finite extension $K|F$ is finite, since it is the splitting field of the minimal polynomials of a set of generators of $K|F$, and there are finite sets of generators.
Nov
20
awarded  Enlightened
Nov
20
awarded  Nice Answer
Nov
13
answered Basic algebraic geometry question (confused about conventions)
Nov
10
comment Reproducing kernel
What do you mean by "matrix valued kernel"? If $H$ is a Hilbert space functional over some set $X$, then it determines a unique kernel function $K:X\times X\rightarrow\mathbb{R}$ (I'm assuming $H$ is a real vector space). Vice versa every kernel function gives a functional Hilbert space. But kernel functions are always real-valued here.
Nov
3
answered Sppliting field and Galois theory and its Automorphism group
Oct
27
answered Example of a normal extension.
Oct
27
answered Finite modules over (infinite) commutative rings