4
votes
Every finite integral domain is a field (why is it commutative?)
An integral domain is commutative by definition, or at least the standard definition. (The term "abelian" is specific to groups, not ring multiplication.) You're thinking of what's generally ...
2
votes
What is the Conway polynomial for $GF(2^{256})$?
When I saw how low the one Buchanan found was ($X^{256} + X^{10} + \dots$), and thought about that possibly being an upper-bound (if that one happens to be primitive, then worst-case $2^{10}$ attempts ...
1
vote
Evaluating a quadratic character sum.
One way (rather standard, I think):
Consider the equation
$$x^2-D=y^2\qquad(*)$$
with $D$ as in your post and $x,y$ ranging over $\mathbf{F}_q$. Denote by $N(D)$ the number of solutions $(x,y)$ of $(*)...
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