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 Yearling
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Mar
3
awarded  Yearling
Feb
11
comment Maximum Modulus Exercise
I don't know. One might be able to prove something like that in general, but first one would have to say what exactly is meant by "farthest away from the zeroes".
Feb
1
comment Injectivity and localisation in Rings
As to why $f: \operatorname{Spec} B \to \operatorname{Spec} A$ is dominant, consider $s \in A$ with $f(\operatorname{Spec} B) \subseteq V(s)$. Then $\varphi(s)$ is contained in every prime ideal of $B$, hence nilpotent in $B$, hence nilpotent in $A$.
Jan
14
comment Does commutativity imply Associativity?
Well, associativity and commutativity are properties of maps $X\times X \to X$ for a set $X$. In other words, such a map takes two elements as an "input" and returns a single element. In my example, the set under consideration is the set of integers and the map sends each pair of integers $(x,y)$ to $xy+1$. Commutativity means $xy + 1 = yx + 1$ for all $x$ and $y$, which is satisfied. Associativity would mean $x(yz+1) + 1 = (xy+1)z+1$ for all $x$, $y$ and $z$, but it's easy to find examples where this equation does not hold, so the operation is not associative.
Dec
16
comment If $F$ is finite over $L_1$ and $L_2$, is it also finite over $L_1 \cap L_2$?
I have thought about it, but I couldn't prove it nor find a counterexample.
Dec
16
asked If $F$ is finite over $L_1$ and $L_2$, is it also finite over $L_1 \cap L_2$?
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Jul
27
awarded  Good Answer
May
23
comment Finding inverse of polynomial in a field
@miloszmaki: Remember that the polynomials have coefficients in $\mathbb F_3$, so that $4 = 1$ and $1/2 = 2$ and $-1 = 2$. Thus the solution you propose is actually the same as mine.
Apr
15
answered Isomorphism of intervals of a distributive lattice
Mar
3
awarded  Yearling
Feb
13
awarded  Necromancer
Jan
17
reviewed Approve The random walk of two drunks
Nov
29
revised Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
replaced an unclear step in the argument
Nov
29
revised Evaluation of a product of sines
fixed typo + formatting
Sep
10
answered Direct way to show: $\operatorname{Spec}(A)$ is $T_1$ $\Rightarrow$ $\operatorname{Spec}(A)$ is Hausdorff
Aug
15
awarded  Nice Answer
Jun
25
comment Number of prime ideals of a ring
Thanks, I've corrected that.
Jun
25
revised Number of prime ideals of a ring
replace $\sigma(m)$ with $\tau(m)$