So, this is from "a classical introduction to modern number theory" and I just don't know why the union is an ideal. It is not "easy to see". This is supposed to be a principle ideal domain also.
#Why is this set an Ideal
long gate
ancient stratusBOT
So, this is from "a classical introduction to modern number theory" and I just d...
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quick viper
So, this is from "a classical introduction to modern number theory" and I just d...
Well, every a_i is a subset of a_(i + 1), so I is just, well, a_infinity, as it were.
long gate
Well, every a\_i is a subset of a\_(i + 1), so I is just, well, a\_infinity, as ...
But why is this an ideal?
quick viper
But why is this an ideal?
...because every a_i is an ideal, and I just equals one of those.
long gate
...because every `a_i` is an ideal, and I just equals one of those.
Doesn't it equal to a_1?
quick viper
Doesn't it equal to a_1?
Not necessarily.
long gate
...because every `a_i` is an ideal, and I just equals one of those.
Why is it equal to one of those?
quick viper
Hold on.
quick viper
Why is it equal to one of those?
Because. Again. a_i is a subset of a_(i + 1), which means their union is just a_(i + 1).
long gate
Because. Again. `a_i` is a subset of `a_(i + 1)`, which means their union is jus...
Right... But it is an infinite union
quick viper
Right... But it is an infinite union
Do you agree that every finite union across a_i is equivalent to one such a?
quick viper
You also have the condition with the k and l there.
long gate
You also have the condition with the k and l there.
That is what we are trying to prove though?
quick viper
Ah. So then the hypothesis is just the part about the ascending chain of ideals.
long gate
Ah. So then the hypothesis is just the part about the ascending chain of ideals.
Yes... otherwise it is trivial
quick viper
Is it possible for I to not be an ideal?
long gate
Is it possible for I to *not* be an ideal?
The reason I posted this question, is because I cannot see why it should be (i.e. I have already tried to see what happens if it isn't)
cyan mirage
The reason I posted this question, is because I cannot see why it should be (i.e...
Have you gone through checking the 3 requirements for a set to be an ideal?
long gate
Have you gone through checking the 3 requirements for a set to be an ideal?
How is it $I$ an additive subgroup?
zenith mountainBOT
Addict
cyan mirage
How is it $I$ an additive subgroup?
If $f,g\in I$ then $f\in (a_i)$ and $g\in (a_j)$ for indices $i,j$
zenith mountainBOT
Omegabet_
cyan mirage
if $i=j$, trivial.
so, WLOG, assuming $j>i$.
By the chain condition, $f\in (a_i)\subseteq (a_j)$, so both $f$ and $g$ are in $(a_j)$, hence $f+g\in (a_j)\subseteq I$
zenith mountainBOT
Omegabet_
long gate
if $i=j$, trivial.
so, WLOG, assuming $j>i$.
By the chain condition, $f\in (a_i...
Actually yeh, I was unclear of the definitions of ideals, the authors assumes the knowledge. Next time I will make sure I understand everything
cyan mirage
That'd....... be useful.
runic flaxBOT
@long gate