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Apr
18
comment Localization Preserves Euclidean Domains
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
Apr
18
revised Localization Preserves Euclidean Domains
added 557 characters in body
Apr
18
answered Localization Preserves Euclidean Domains
Apr
18
revised Hall subgroup property
Markup / LaTeX fixed
Apr
16
answered Blow-up and resolving a singularity
Apr
16
comment Hypersurfaces have no embedded points (Vakil 5.5.I)
It is of height one because it is principal and prime because it is the preimage of the prime ideal $P$.
Apr
16
comment Understanding functions in $\mathbb{P}^1$
The divisor of this function is $[0:1]-[1:0]$, because it has a (single) zero at $[0:1]$ and a (single) pole at $[1:0]$, as you pointed out correctly.
Apr
16
answered Hypersurfaces have no embedded points (Vakil 5.5.I)
Apr
15
comment Intersection of hypersurfaces in the projective space
This is true if you are thinking clasically, not scheme-theoretically. Let $\mathfrak m$ be the maximal ideal of your point $x$ and let $\mathfrak m^k = (f_1,\ldots,f_m)$. The $f_i$ are homogeneous of degree $k$ and each of them defines a hypersurface $Z_i$. Then, $Z_1\cap\cdots\cap Z_m=Z(f_1,\ldots,f_m)=Z(\mathfrak m^k)=Z(\mathfrak m)=\{x\}$.
Apr
14
comment Maximum matchings in a bipartite graph
But you said "$S$ is a set of vertices $S\subset X$".
Apr
14
comment Maximum matchings in a bipartite graph
This is precisely the Hall condition.
Apr
13
comment the intersection of an empty family of sets; what's wrong with this proof?
Your mistake is when you say "then $x\in A$ for some $A\in S$". There is no such $A$ if $S=\emptyset$. In fact, when $S$ is the empty set, then $\forall A\in S\colon x\in A$ is true for every set $x$. A good way to check for mistakes is by going through your proof and substituting $S$ for $\emptyset$ all the time and see what happens.
Apr
13
comment Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$
To be frank, I used a computer algebra system simply to confirm my suspicion. But the general method is outlined on wikipedia. It's a bit tricky.
Apr
13
answered Normal closure of $\mathbb{Q}(\sqrt{11+3\sqrt{13}})$ over $\mathbb{Q}$
Apr
13
answered Morphism whose fibers are finite and reduced is unramified
Apr
13
revised Prove that the determinant of an invertible matrix $A$ is equal to $±1$ when all of the entries of $A$ and $A^{−1}$ are integers.
LaTeX
Apr
13
answered Prove that the determinant of an invertible matrix $A$ is equal to $±1$ when all of the entries of $A$ and $A^{−1}$ are integers.
Apr
12
revised Finding the Dimension of a given space V
added 13 characters in body
Apr
12
revised Finding the Dimension of a given space V
added 453 characters in body
Apr
12
answered Finding the Dimension of a given space V