# Ayman Hourieh

Dublin, Ireland

 2d awarded Great Answer Apr 18 comment Atiyah-Macdonald, Exercise 4.6Related Apr 2 awarded Guru Mar 18 comment Is the set $\{\big(x,\sin(1/x)\big):x\ne 0 \}$ connected in usual metric of $\mathbb R^2$?Well, $\mathbb R = (-\infty, 0) \cup [0, \infty)$. These intervals are connected and disjoint. You need more in order to show that each subset is a distinct component. Mar 17 comment Is the set $\{\big(x,\sin(1/x)\big):x\ne 0 \}$ connected in usual metric of $\mathbb R^2$?You also need to show that each of $f(\mathbb R^+)$ and $f(\mathbb R^-)$ is open (or closed). Mar 14 comment How to find the elements of a finite field?The statement in your last paragraph is true. It's usually a theorem proved soon after introducing polynomial rings over fields. What reference are you using? Feb 28 comment Minimal prime ideals of $\mathcal O_{X,x}$ correspond to irreducible components of $X$ containing $x$Let $\{U_j\}$ be the irreducible components of $U$ containing $x$. The map $U_j \mapsto \overline U_j^X$ gives a bijection from $\{U_j\}$ to $\{X_i\}$ whose inverse is $X_i \mapsto U \cap X_i$. This is an exercise in point-set topology. Feb 25 comment Compactness of $Y$ implies compactness of $X$I half-typed an answer but euouae posted one before me. What's wrong with that proof? Feb 21 comment Compactness of $Y$ implies compactness of $X$This may be easier to prove with the formulation of compactness that uses the finite intersection property. Are you interested in such a proof? Feb 17 revised Subgroups containing kernel of group morphism to an abelian group are normal.added 14 characters in body Feb 16 comment Field theory extensions.Proof$K(\alpha_1, \ldots, \alpha_k)$ is the smallest subfield of $L$ that contains both $K$ and $\{\alpha_1, \ldots, \alpha_k\}$. Feb 16 reviewed Reject suggested edit on What's the point in being a "skeptical" learner Feb 15 awarded Nice Answer Feb 14 comment cofinite topologyYou don't have to look at the real line specifically. If a space has the cofinite topology, then it's compact. If the space is infinite, then it's also connected. Try to prove this. Feb 13 reviewed Approve suggested edit on How do I use homomorphism theorem to show the assertion? Feb 12 comment Why is a discrete algebraic subset of $K^n$ finite?@Drike I don't think so. $\operatorname{Spec} R$ is always quasi-compact, regardless of whether $R$ is Noetherian or not. Feb 11 comment Let $F : X → X$ be continuous. Prove that the set $\{x ∈ X : F(x) = x\}$ of fixed points of F is closed in XHint: A space $X$ is Hausdorff iff the diagonal is closed in $X \times X$. Feb 10 comment Hartshorne Exercise II.2.18(d)@Manos No. $\mathfrak p$ is a point in $\Spec A$. Think of $B$ as an $A$-algebra, and localize it at the prime ideal $\mathfrak p$ of $A$. This is a valid operation that gives us the homomorphism $\varphi_\mathfrak p$. Remember that $f^\#$ is a morphism of sheaves on $\Spec A$. This is why we consider points in $\Spec A$. Feb 10 comment Hartshorne Exercise II.2.18(d)@Manos If we consider $B$ as an $A$-algebra via $\varphi$, then the notation $B_\mathfrak p$ makes sense. Feb 10 reviewed Approve suggested edit on Let $f$ be injective and discontinuous at some point $c$. Can its inverse be continuous?