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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
 

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