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May
16
awarded  Explainer
May
16
comment In a commutative ring with identity, if $p$ is irreducible, is ($p$) a maximal ideal?
In $\mathbb{Z}[x]$ the polynomial $x$ is irreducible but $\mathbb{Z}[x]/(x)\cong \mathbb{Z}$ is not a field.
May
16
answered if $rank{(A - \lambda I)^k} = rank{(B - \lambda I)^k}$ then $A$ is similar $B$
May
16
revised if $rank{(A - \lambda I)^k} = rank{(B - \lambda I)^k}$ then $A$ is similar $B$
added 1 character in body
May
11
answered 'Inverse' property of a group and the special case that makes a group an Abelian group
May
11
revised 'Inverse' property of a group and the special case that makes a group an Abelian group
Tex
May
10
answered Quotient of ideals of the ring of rational numbers with denominator prime to p.
Apr
30
comment What are the subfields of $\mathbb{Q}(\sqrt{3},\sqrt[7]{5})$?
@pretzelman : because a primitive root of unity of order 7 , is algebraic of degree 6.
Apr
27
comment Show that $B^{(1)} = 0$ and $B^{(2)}$ have basis $\{[x_i,x_j]; i>j\}$
Since $deg([x_{i_1},...,x_{i_{k_1}}][x_{i_{k_1}+1},...,x_{i_{k_2}}]...[x_{i_{k_l}+1},.‌​..,x_{i_n}])=n$, if we want the degree of a product of commutators to be 2, then there must be only one factor in the product, and it can has only two variables, namely $[x_i,x_j]$.
Apr
25
answered Show that $B^{(1)} = 0$ and $B^{(2)}$ have basis $\{[x_i,x_j]; i>j\}$
Apr
25
revised Maximal ideals and prime ideals of $\mathbb{Z}/2 \times \mathbb{Z}/2$?
tex
Apr
10
answered Does Greens Theorem apply to the annulus?
Apr
10
comment Rigorously prove $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2) (\sqrt3)$
Can you show that one of these fields is contained in the other?
Apr
10
comment Applying Chain rule to $z = z(u, v) = f(x(u, v), y(u, v))$.
You should give a better title to your question, than just its first sentence.
Apr
10
revised Applying Chain rule to $z = z(u, v) = f(x(u, v), y(u, v))$.
tex
Apr
10
answered A clique in a tree decomposition is contained in a bag
Apr
10
comment number of roots of unity which satisfy a given polynomial
This is a nice approach. The problem is that $u(t),v(t)$ are not polynomials. If on the other hand we multiply by $(1+t^2)^{\deg(f)}$, then they become polynomials but with degree $2\deg(f)$. Is there a way to bound $\deg(gcd(u,v))$ in this case?
Apr
10
awarded  Curious
Apr
9
asked number of roots of unity which satisfy a given polynomial
Apr
9
comment Number rings as free module over base ring
@Asvin : You are right, it should be $\mathcal{O}_L$ - fixed it. In case it is free, its rank is $[L:K]$. You can see this by noting that $[\mathcal{O}_L:\mathcal{O}_K][\mathcal{O}_K:\mathbb{Z}]=[\mathcal{O}_L:\mathbb{‌​Z}]$.