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Jun
18
awarded  Fanatic
Jun
2
revised Polynomial in several variables over $GF(2)$
Improved legibility.
Jun
2
suggested approved edit on Polynomial in several variables over $GF(2)$
Jun
2
revised If I invert the argument, should I invert the constants in the equation?
Made it easier to read.
Jun
2
suggested approved edit on If I invert the argument, should I invert the constants in the equation?
May
27
comment Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
$-1$ seems to work...
May
14
suggested rejected edit on complex nos in ellipse.
May
12
comment Quadratic solutions puzzle
(-1/2, -1/2) is a solution, but it is also said in the constraints that $a \neq b$.
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, thanks! It is just hard to wrap my head around. A lot of things in abstract algebra are requiring some serious rewiring of my brain. sigh
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, so then we can construct any numerator as a unique combination of primes (well known property of primes) and choose a specific bottom element as a particular unit so this would imply that R is a UFD then?
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: When you say $2$ is a unit, are you referring to the $p \times 2^{-1}$ is unnecessary just as $-1$ is unnecessary in prime factorization?
May
11
asked Confused about the prime elements of a ring.
May
10
awarded  Yearling
May
10
accepted Need help determining what this center is isomorphic to
May
10
comment Need help determining what this center is isomorphic to
Oh, whoops my bad, but $Z$ should have order $2$. That was me slipping up there, I didn't mean to write that, just mixed things around sorry about that. Thanks! (Modified original comment.)
May
10
comment Need help determining what this center is isomorphic to
Ok, so I think the problem is I didn't know about p-group. In this case the argument should go: Since it is non-trivial, we know that the order is $2$, $4$ or $8$, but G is not abelian so the order cannot be $8$ and $2$ is guaranteed cyclic hence $Z$ has order $2$ and $|G/Z| = 4$
May
10
asked Need help determining what this center is isomorphic to
May
8
accepted Help show the following isomorphism cannot exist.
May
8
comment Help show the following isomorphism cannot exist.
Ok, so I think I get it, I made a small typo in my question, but was able to follow through and get $v^2 - v + 1 = 0$. But still, $x^2 - x + 1 \not\in \langle g(x) \rangle$. Thanks!
May
8
revised Help show the following isomorphism cannot exist.
edited body