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12:00 AM
I never did the symmetric polynomial approach, Artin's is much easier. I don't know that it would be true that every quantic polynomial's galois group would be S_5
quintic*
 
For $p,q,r,s$ in some field, how would I prove from the field axioms that $\dfrac{p}{q}\cdot\dfrac{r}{s}=\dfrac{pr}{qs}$, i.e., $(pq^{-1})(rs^{-1})=(pr)(qs)^{-1}$?
 
every irreducible one perhaps
 
($q,s\neq0$ here)
 
Is there something I can do to make latex commands show up in these chat rooms. i keep getting the source language instead of the rendered output
 
@DavidReed Look up "ChatJax"
 
12:02 AM
ty
 
no, look at the link to LaTeX in chat in the upper right @DavidReed
 
@TedShifrin The top search result (for me at least) gives the same link, so ¯\_(ツ)_/¯
 
oh, I guess it depends where you search, Legion
 
But yeah, would you have any idea how I'd solve my current problem?
The equation seems pretty basic
 
@LegionMammal978 first show that $b^{-1}a^{-1}$ is an inverse of $ab$ (this works in any group)
the rest follows from commutativity and associativity
I guess you also need to use the uniqueness of inverses
but I think if you use the notation $a^{-1}$ you already asumme that
 
12:07 AM
This seems sort of silly to me. Multiplication is commutative and associative ... so ... done.
 
is this the regex for this language: x starts with 0 and does not contain substring 101
 
I'm just not sure how to get from, as @Mathein said, from $aa^{-1}=1$ to $a^{-1}b^{-1}=(ab)^{-1}$
 
0*(1*000*)*1*0
 
@LegionMammal978 what happens if you multiply $ab$ and $a^{-1}b^{-1}$?
 
Well, what is $(ab)(a^{-1}b^{-1}) = a(a^{-1}b^{-1})b$?
 
12:10 AM
Assuming you've defined a field to be a commutative division ring via the notion of group of units, uniqueness is implied. mathein's approach is the correct way.
 
Ah, I see now
Lots of commutativity and associativity
 
That's what I said :P
Granted, I was a bit short.
 
@Theo The string 01 is contained in the language but does not match your regex
 
@LegionMammal978 why not? since (1*000) does not have to be present, it can show up zero times
there we are left with 0*1*0
 
@Theo It does not contain the final 0
 
12:15 AM
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
 
I just proved that the set of all zero divisors, along with $0$, in a commutative ring forms an ideal. Is this right, or did I make a mistake?
 
this is wrong
consider $\Bbb Z/(6)$
$3$ and $2$ are zero-divisors
 
products work, but not sums :)
 
but $3-2$ is not
 
Ah...I see...
 
12:21 AM
@MatheinBoulomenos Oh ok. I don't believe I've ever encountered that terminology. Thank you for letting me know.
 
@LegionMammal978 what about this? 0*(1*000*)*1*0*
 
@Theo That matches the empty string, which is not in the language
 
Okay...If $d$ is a zero divisor, then will the principal ideal $(d)$ consist entirely of zero divisors?
 
@LegionMammal978 0*(1*000*)*1
that should do it i believe
 
12:26 AM
@user193319 Good other rules to know for ideals: $(a) = (b)$ iff $a,b$ are associates and $(d) = R$ iff $d$ is a unit
 
@Theo Doesn't match 0.
 
the implication $(a) = (b)$ implies $a$ and $b$ are associates only works in integral domains iirc
 
iirc?
 
if I recall correctly
 
ah
You may be right, when I went through algebra the notion of an ideal was only defined for elements of integral domains
If that's true then I stand corrected.
 
12:30 AM
@LegionMammal978 this? 0*(1*000*)*0
@LegionMammal978 nvm doesn't cover 01
@LegionMammal978 is this possible? 0(0*)(1*000*)*
 
@Theo Doesn't match 01.
 
why not? 0(no 0 after)(one 1, no 000)
 
That would be the regex 00*(1*(000)*)*
And that doesn't match 010
 
is it possible to do: 0(1*0*)*
that way 101 won't occur, and 01, or 0000, or 010 still occurs
 
@Theo That matches 0101.
 
12:35 AM
oh right
okay im a bit lost, I know 0*(1*000*)*1*0* will make it so 101 does not occur in substring
but how do i avoid an empty string
and make sure it starts with 0
 
gimme a minute
 
hey
anyone like graph theroy?
anyone still here? :-o
 
12:56 AM
I haven't gone through graph theory in nearly a decade
My guess is if nobody here is up for it, comp sci ppl use it a lot, you may have luck in stack overflow chat
 
@Theo Okay, finally got 0(0|11*00)*(ε|11*|11*0)
Note that the last union operation includes the empty string
 
do we need the empty string there?
 
Yes, in case the string doesn't end with 11* or 11*0
You can omit the epsilon if you'd like
 
Thanks @DavidReed :)
 
Or expand out the regex into 0(0|11*00)*|0(0|11*00)*11*|0(0|11*00)*11*0
 
1:08 AM
ahh gotcha, thanks so much!
 
Yeah, ended up having to design an FSM and do manual state reduction
 
@alan2here Have to warn you though...they can be mean. Typical comp-sci nerd I know everything narcissism
 
In computer-regex, it could also be written 0(0|1+00)*(1+0?)?
 
and im assuming this is much more difficult to get a regex for
either all the symbols in odd positions within w are
0's, or all the symbols in even positions within w are 0's, or both.
 
@Theo gimme a few minutes
Is w 0- or 1-indexed?
Although I guess it doesn't really matter
 
1:21 AM
w $\in$ {0, 1}
 
As in, is the first position odd or even
 
i'm not too sure what you mean by 0- or 1-indexed
first position is odd
 
Okay
This is definitely possible
 
@DavidReed Including relevant greek mythology?
 
@LegionMammal978 how?
 
1:33 AM
@Theo I've already finished designing the state machine
 
@LegionMammal978 i tried doing (01)*(10)*(0)* but that leaves for an empty string
 
@Theo An empty string should be allowed
But that particular regex matches stuff like 0110
 
@Narcissusjewel Intellectual narcissism. Greek would be closer to somatic narcissism I believe.
 
@DavidReed I was just trying to make myself relevant, you know how we are ;).
 
hahaha.Well now you're relevant! Tell me what motivated that name.
 
1:43 AM
@LegionMammal978 how did you do it then?
 
@Theo As I said, FSM state reduction
I have a regex now, currently simplifying it as much as I can
 
okay
 
@Theo Okay, got 0*|1|01(01)*(ε|0)|10(10)*(ε|1)
It has 4 cases: 0000..., 1, 0101..., and 1010...
 
is 1 a case
?
I thought 0 had to be in the string
whether in even position, odd, or both
 
Whoops, slightly misinterpreted the language specification
Just remove that case, then
Although I suppose "all even positions are 0" could be considered vacuously true due to there being no even positions
Same with the empty string (to remove that you'd change the first case to 00*)
 
1:58 AM
00*|1|01(01)*(ε|0)|10(10)*
we'd end up with this then
 

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