marlu
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 Mar3 awarded Yearling Feb11 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". Feb1 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 $A$. Jan14 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. Dec16 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. Dec16 asked If $F$ is finite over $L_1$ and $L_2$, is it also finite over $L_1 \cap L_2$? Sep30 awarded Explainer Sep24 awarded Autobiographer Jul27 awarded Good Answer May23 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. Apr15 answered Isomorphism of intervals of a distributive lattice Mar3 awarded Yearling Feb13 awarded Necromancer Jan17 reviewed Approve The random walk of two drunks Nov29 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 Nov29 revised Evaluation of a product of sines fixed typo + formatting Sep10 answered Direct way to show: $\operatorname{Spec}(A)$ is $T_1$ $\Rightarrow$ $\operatorname{Spec}(A)$ is Hausdorff Aug15 awarded Nice Answer Jun25 comment Number of prime ideals of a ring Thanks, I've corrected that. Jun25 revised Number of prime ideals of a ring replace $\sigma(m)$ with $\tau(m)$