Graduate student mostly dabbling in algebraic geometry.

2d
comment What is a basis and dimension of $span\{I,M,M^2,...\}$ where $I$ is the identity matrix and $M$ is invertible squared matrix?
@Omnomnom I think you mean $\{I,M,\ldots,M^{n-1}\}$ is a basis if the dimension of the span is $n$?
Feb
9
comment Roots of $x^p + x + [\alpha]_p \in \mathbb{Z}_p[x]$
As suggested, this works for all $\alpha\in\Bbb{Z}$ if $p\neq2$.
Feb
9
comment What is a basis and dimension of $span\{I,M,M^2,...\}$ where $I$ is the identity matrix and $M$ is invertible squared matrix?
1: The matrix you give is not square. 2: A basis of your $\text{span}$ will consist of matrices, not of (column) vectors. 3: I can't make sense of your last sentence about representing $I$. Would you care to expand?
Feb
9
answered What is a basis and dimension of $span\{I,M,M^2,...\}$ where $I$ is the identity matrix and $M$ is invertible squared matrix?
Feb
9
answered Roots of $x^p + x + [\alpha]_p \in \mathbb{Z}_p[x]$
Feb
3
revised Does this base change yield another dominant morphism?
added 438 characters in body
Feb
2
revised Does this base change yield another dominant morphism?
edited title
Feb
2
asked Does this base change yield another dominant morphism?
Jan
11
answered Equation in the complex plane $8z=i|z|^3\bar{z}$?
Dec
25
comment The degree of field extension
Would anyone care to motivate the sudden downvote?
Dec
25
comment How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$
@Vim: You are absolutely right. The hint is meant just to get across the 'essence' of the problem.
Dec
25
answered How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$
Dec
25
revised How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$
added 5 characters in body; edited tags
Dec
25
comment How to solve $\sqrt {1+\sqrt {4+\sqrt {16+\sqrt {64+\sqrt {256\ldots }}}}}$
What are your thoughts on the problem? What have you tried, and where did you get stuck?
Dec
19
comment If $\{x_1, x_2\}$ are orthogonal vectors in $\mathbb R^3$, how to prove that we can always find the third vector so that they become orthogonal basis?
Assuming that $x_1$ and $x_2$ are nonzero...
Dec
18
accepted Editing (and creating) seemingly useless tags
Dec
17
answered Use congruences to factor $n=87463$ (Fermat's Factorization?)
Dec
17
comment Use congruences to factor $n=87463$ (Fermat's Factorization?)
@Jake please use MathJax to typeset math.
Dec
17
comment Use congruences to factor $n=87463$ (Fermat's Factorization?)
@David The first few equations are not congruences.
Dec
17
revised Use congruences to factor $n=87463$ (Fermat's Factorization?)
added 129 characters in body; edited tags; edited title
1 2 3 4 5