- #1
Brimley
- 77
- 0
Hello PhysicsForums!
I had been reading up on a congruence classes when I came across an example that miffed me. A lot of information was given but I can't seem to make sense of it: Here it is:
Assume [itex]\lambda = (3+\sqrt{-3})/2 \in \mathbb{Q}[\sqrt{3}][/itex]. We know that [itex]\lambda[/itex] is prime and that there are three congruence classes [itex] (mod \lambda)[/itex], one with 0, one with 1, and the last with -1.
The following three relationships exist:
A. If [itex]x \equiv 1 (mod \lambda)[/itex], then [itex]x^3 \equiv 1 (mod \lambda)^3[/itex].
B. If [itex]x \equiv -1 (mod \lambda)[/itex], then [itex]x^3 \equiv -1 (mod \lambda)^3[/itex].
C. If [itex]x \equiv 0 (mod \lambda)[/itex], then [itex]x^3 \equiv 0 (mod \lambda)^3[/itex].
Can someone help explain to me why these three relationships exist? I don't know if factoring [itex]x^3 - 1[/itex] in [itex]\mathbb{Q}[\sqrt{d}][/itex] completely into linear factors can help, but its an idea.
I had been reading up on a congruence classes when I came across an example that miffed me. A lot of information was given but I can't seem to make sense of it: Here it is:
Assume [itex]\lambda = (3+\sqrt{-3})/2 \in \mathbb{Q}[\sqrt{3}][/itex]. We know that [itex]\lambda[/itex] is prime and that there are three congruence classes [itex] (mod \lambda)[/itex], one with 0, one with 1, and the last with -1.
The following three relationships exist:
A. If [itex]x \equiv 1 (mod \lambda)[/itex], then [itex]x^3 \equiv 1 (mod \lambda)^3[/itex].
B. If [itex]x \equiv -1 (mod \lambda)[/itex], then [itex]x^3 \equiv -1 (mod \lambda)^3[/itex].
C. If [itex]x \equiv 0 (mod \lambda)[/itex], then [itex]x^3 \equiv 0 (mod \lambda)^3[/itex].
Can someone help explain to me why these three relationships exist? I don't know if factoring [itex]x^3 - 1[/itex] in [itex]\mathbb{Q}[\sqrt{d}][/itex] completely into linear factors can help, but its an idea.