Reputation
505
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
2 13
Impact
~4k people reached

Aug
29
accepted Proof of Chow's lemma in EGAII
Aug
29
comment Proof of Chow's lemma in EGAII
Ah yes. I see. This then works for the full generality that EGA claims as it has enough assumptions to keep scheme-theoretic closures to behave well.
Aug
29
comment Proof of Chow's lemma in EGAII
This blasted scheme vs prescheme business of EGA confused me. I see the statement of Chow's lemma claimed in EGA holds. I am stlil curious however if the proof via reducing to irreducible case is salvageable
Aug
29
comment Proof of Chow's lemma in EGAII
The stacks project basically assumes quasiseparted in addition to the weaker hypothesis (if I'm not mistaken), which still is not the generality EGA claims. Is the EGA proof salvageable even in the noetherian case?
Aug
28
revised Proof of Chow's lemma in EGAII
fixed small error
Aug
28
asked Proof of Chow's lemma in EGAII
Apr
29
comment Can you find the lighter and heavier marbles?
So presumably you can't tell if a bag is heavier than others by any method but the scale. If that is the case, if there are more than 2 bags, this is impossible.
Apr
6
comment How do I prove cardinality is well-defined?
@DanielEscudero Another thing I forgot to mention about your response is that it should probably be checked that $I_n\not\sim I_m$ if $n\neq m$. That is implicit in what your method needs for well-definedess (but not what was needed in user45150's original question)
Apr
6
comment How do I prove cardinality is well-defined?
@user42912 I think what is important to remember with regards to Daniel Escudero's definition is that it is a way to associate a size, called the cardinality, to every set $X$. For this you must check that $\sim$ is an equivalence relation because if you associate the number $n$ to $X$ and $m$ to $Y$ and $X\sim Y$ you would like to be able to conclude that $n=m$.
Apr
6
comment How do I prove cardinality is well-defined?
@DanielEscudero The discussion you give above may be a little confusing, because you do not mention what $\text{card}(X)$ actually is. For instance what happens if $X$ is infinite. I agree that the conflation of the definition of cardinality that you give and the one in the original question may be the cause of the confusion. The reason I did not mention this point of view in my comments was that it was not mentioned in the question and I did not want to add confusion, but since it is brought up we may as well develop it.
Apr
4
comment Understanding algebraic closure
On the $[K:F_p]$ point. Notice that $K$ must be countably infinite, and since $F_p$ is finite $[K:F_p]$ must be the cardinality of the natural numbers
Apr
4
comment How do I prove cardinality is well-defined?
The notion is well-defined already, even before you prove it is an equivalence relation. There is nothing to check. This is because the definition: "there exists a bijection" doesn't depend any choice or anything, either there is one or there isn't
Apr
4
comment How do I prove cardinality is well-defined?
I guess I am confused. $\text{card}(X)=\text{card}(Y)$ if and only if there exists a bijection between $X$ and $Y$. This is a legitimate definition. You don't pick a representative for an equivalence set or any other of the usual well-definedness problems.
Apr
4
comment How do I prove cardinality is well-defined?
What do you mean by welldefined? Do wish to show that it is an equivalence relation?
Mar
26
comment Ring homomorphisms on the set of rationals that coincide on integers
Taking $1$ to $1$ is in the definition of ring homomorphism. See for instance:en.wikipedia.org/wiki/Ring_homomorphism, it seems to be standard. Perhaps you are trying to use another definition and that is the confusion?
Mar
25
comment Ring homomorphisms on the set of rationals that coincide on integers
Ring homomorphism by definition takes $1$ to $1$
Mar
24
comment Ring homomorphisms on the set of rationals that coincide on integers
If $f$ is not the zero map, the image of $f$ is a copy of $\mathbb{Q}$. Note that $f(n)=nf(1)=n\cdot 1=n$. How can you use that the image of $f$ is isomorphic to $\mathbb{Q}$ and the fact that $n$ is in the image?
Mar
23
comment Ring homomorphisms on the set of rationals that coincide on integers
Then use Josh Keneda's comment to allow you to divide by $n$.
Mar
21
comment Ring homomorphisms on the set of rationals that coincide on integers
To add to my comment. What happens if you multiply $f(1/n)$ and $g(1/n)$ by $n$?
Mar
21
comment Ring homomorphisms on the set of rationals that coincide on integers
Notice that if you add $p/q$ to itself $q$ times you get $p$. How does this help?