All Questions
7
questions
1
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Homogeneous polynomial of degree $n$ in $n^2$ indeterminates that cannot be the sum of the terms of degree $n$ in $P_1Q_1 + ... + P_k Q_k$
This is a "difficult" exercise taken from Bourbaki's "Commutative Algebra". I have no idea on how to solve it, nor tackle it.
Let $K$ be a finite field. Prove that for all $k \in \...
- 2,011
2
votes
1
answer
147
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Maximal ideals in $\mathbb{F}_q[x,y]$ [closed]
Let $p \in \mathbb{P}$ be a prime and suppose that an integer $e > 1$ is given such that the polynomial $s_e = 1 + x + \dots + x^{e-1}$ is irreducible in $\mathbb{F}_p[x]$.
My question is the ...
- 380
3
votes
1
answer
201
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Is there a principal maximal ideal in $\mathbb F_q[X,Y]$? [duplicate]
Given an infinite field $K$, one can prove that any maximal ideal of $K[X,Y]$ can't be principal. In fact, every non-principal prime ideal is a maximal ideal, and can be generated by two polynomials.
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- 5,571
1
vote
1
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203
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Degrees in Monomials
I am looking over the Joux-Vitse algorithm paper whereby they present an algorithm that seems to outperform exhaustive search and some state-of-the-art algorithms. However, it only works with ...
3
votes
0
answers
68
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Is there a way of multiplying a polynomial in X &Y with fractional powers with another such polynomial to get polynomial in X &Y with integer powers?
I want to take something like $X^{m/k}+X^{(m-1)/k}+Y^{1/k}$ and multiply it by another polynomial in $X^{1/k}$ and $Y^{1/k}$ to get a polynomial without fractional powers, ideally irreducible in $K[x,...
- 89
4
votes
2
answers
140
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Polynomials over $\mathbb{F}_2$ with certain values in $\mathbb{F}_4$
Let $\mathbb{F}_4=\{0,1,u,u^2\}$ be the field with $4$ elements. Is there a polynomial $p \in \mathbb{F}_2[x,y]$ with the following property?
(1) For $r,s \in \mathbb{F}_4$, we have $p(r,s)=u \...
- 165k
0
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0
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138
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Multivariable irreducible polynomials over finite fields
It is not difficult to prove the following result, and it seems that it should be already proved. I would appreciate it if someone offer me some reference to it.
For any $f(x_1,\dots, x_n)=\sum c_{...
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