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20:09
Did someone say "forsake analysis completely"? I have been summoned
20:24
@Fargle :0
Is it that you're planning on it?
Nah, just memeing, lol. I'm disappointed in how weak I feel in the area.
@AbdullahUYU That is a beautiful letter A you have there.
@JasperLoy I remember drawing it in Inkscape with my mouse for having an icon :)
@AbdullahUYU I drew my picture using Paint in Windows.
afair, it has a calligraphy tool.
cool :D
20:28
Ah I see
@JasperLoy Actually, I lost the original svg file quite a while ago. But it remained in google+
@AbdullahUYU To delete all the pics you posted on google ever, you need to go to get.google.com/albumarchive while logged into your account.
I checked it out. Thanks.
20:47
does the algebraic closure of $\Bbb Q_p$ have a discrete valuation?
@Daminark
@Leaky beyond my current level of knowledge :/
oh sorry lol
are you interested?
I can tell you how to construct the maximally unramified extension lol
@Daminark or is there anything in particular you would like to know
(given that they are in my knowledge)
Anything you think is worth mentioning
20:54
so let $K/\Bbb Q_p$ be unramified and degree $n$
then the residue fields $k/\Bbb F_p$ is of degree $n$
so $k = \Bbb F_p(\mu_{p^n-1})$
where $\mu_{p^n-1}$ is the set of $(p^n-1)^{st}$ roots of unity
then by Hensel's lemma, $\Bbb Q_p(\mu_{p^n-1}) \subseteq K$
and then my text says, by comparing degree we can obtain $\Bbb Q_p(\mu_{p^n-1}) = K$
(Side note: what's Hensel's lemma?)
@LeakyNun I suppose 0 can be considered a limit ordinal if you remove the condition that a limit ordinal not be 0 :-)
so Hensel's lemma for the $\Bbb Z_p$ case states that if you have a polynomial $f \in \Bbb Z_p[X]$ and $r \in \Bbb F_p$ such that $f(r) \equiv 0 \pmod p$ and $f'(r) \ne 0 \pmod p$ then you can lift $r$ all the way to a solution of the polynomial in $\Bbb Z_p$
an example is finding $\sqrt2$ in $\Bbb Z_7$
because $x^2-2$ has a solution in $\Bbb F_p$
namely, $3$
I ask about 0 and $\aleph_0$ being considered inaccessible cardinals because they arise from the only existential ZFC axioms (empty set and infinity, respectively).
@Daminark am I being unclear lol
21:19
Sorry I was out for a second, this makes sense I think
so $K^{nr} = \varinjlim \Bbb Q_p(\mu_{p^n-1})$
What does " does the algebraic closure of ℚp have a discrete valuation? " mean? Are you asking for any discrete valuation, or that extends the one on $\mathbb{Q}_p$?
the latter
don't think that exists
point being you have no way to assign value "discretely" to $\{p^{1/n}\}$
I can't tell if I'm messing with myself or my text is messing with me
they say that if you have a degree-$n$ extension $L/\Bbb Q_p$, then you define $\|x\|_L = \|N_{L/\Bbb Q_p}(x)\|^{1/n}$
21:28
but that's finite
finite extension is fine
never mind, I see what is wrong with my thinking
I was thinking that the uniformizer in $L$ would be sent to the same norm as the uniformizer in $\Bbb Q_p$
that can be done, depending on how you define discrete valuation
but if you consider $\Bbb Q_p(\sqrt p)$, clearly $\sqrt p$ is sent to $|p|^{1/2}$
if you mean something like surjection $K^* \to \mathbb{Z}$ then no
I know it can be done, but we don't want to do it now
I was just confused
thank you for your attention
21:30
if you mean something like $K^*$ mapping to a discrete subgroup of $\mathbb{R}$, and that you define equivalence of valuations by linear scaling, then yes
yeah
hello, i have a semi norm $N: \mathbb{R}^n\rightarrow \mathbb{R}$ defined by $N(P)= \sum_{i=1}^m |P(x_i)|$ where P is a polynomial of degree less or equal n, what is the condition to obtain that N is a norm
hello, the condition is that N has to be a norm
@Vrouvrou You need to check the definitions of seminorm and norm and see what additional condition needs to be satisfied.
Hello @MikeMiller hope you are well.
21:51
I hope so too.
Ambiguous sentence you made.
N is a norm if N(P)=0 implise that P=0 , and the sum equal 0 if each of them equal zero then P=0 i don't see where is the problem @JasperLoy
what can be the condition?
I don't think $N(P) = \sum_{i=1}^m |P(x_i)|$
what does $N$ depend on ?
yes it is
21:55
implicitly
no it isn't
oh wait lol I think I might have misread
on $x_i$ ?@JasperLoy
$N$ is a semi-norm on what ?
on R_n[x]
21:57
but you wrote $N : \Bbb R^n \to \Bbb R$ just earlier
$\Bbb R^n$ is very much not the same thing as $\Bbb R_n[x]$
no it is a error $N: \mathbb{R}_n[x] \to \mathbb{R}$
so how many seminorms do you have ?
one just N
so how much is $N(1+x)$ ? is it $7$ ?
i don't understand
22:03
you say you have one seminorm
so a function from $\Bbb R_n[x]$ into $\Bbb R$
I am asking you which real number is $N(1+x)$
$N(1+P)= \sum_{i=1}^m |1+P(x_i)|$
what is $m$
what is $x_1$
$x_1,...,x_m$ are m real distinct numbers
@mercio
and what can those numbers be
how many seminorms do you have in total ?
i don't there is no other indications
22:12
what can the condition you have to find be talking about ?
give me an example of an other semi norm u don't see
well there is the seminorm $N(P) = |P(12)| + |P(874)|$
on $\Bbb R_6[x]$
do you think it is a norm ?
just if P is the polynome zero
what
that's not what being a norm is about
besides the "is it a norm ?" has a yes/no answer
either it is yes
either it is no
I haven't even defined $P$ so I don't know what you mean with "it is a norm if $P$ is zero"
really I don't know
P is a polynomial of degree less or equal n
22:16
in my example $n$ is $6$
and it is not that
$P$ is a letter we use in order to define $N$
a letter which refers to a polynomial
if you think $P$ is a polynomial of degree at most $6$ you should at least tell me what it is exactly
 
1 hour later…
23:43
Could someone of you take a look at the edit part of my question: math.stackexchange.com/questions/2884584/… ?
This place looks serious!
@MaryStar can you show $f(-(x,y)) = -f(x,y)$?

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