2d
awarded Great Answer
Apr
18
comment Atiyah-Macdonald, Exercise 4.6
Related
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 topology
You 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 X
Hint: 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?
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