VW-pre-mop-black handout
by Wolstenholme, Nov 25, 2014, 3:39 AM
Like I said in my earlier blog post, I'm gonna start this thing as both a way to practice for TST's and to transition away from contest math to real math. You can find the link here: https://www.dropbox.com/sh/kzf5l6uyzgkk2tj/AADuNIyfzZFYfQ1AbRLXVOhJa/Olympiad%20stuff%20(mostly%20from%20HS)/MOP%202014%20handouts/VW-pre-mop-black.pdf?dl=0. Here it goes:
Well this immediately follows from 
and 
so let's solve those.
Let 
for some 
where 
for all 
Then 
in ![$ \mathbb{Z}_p[x] $](//latex.artofproblemsolving.com/c/7/6/c760f9013d69b137936d845f8cd71f73d9f12974.png)
by the Frobenius endomorphism. And as it's clear that 
for all 
we obtain the desired result.
First I prove the "only if" direction. Let 
be a root of 
in the splitting field of over 
Note that since 
we have that 
and so ![$ \mathbb{F}_p[\alpha] $](//latex.artofproblemsolving.com/9/7/4/97428c0562aa53c0e15c7c5bc93fa8d52f00b389.png)
is a subfield of 
Then by the Tower Law we find that ![$ [\mathbb{F}_{p^r} : \mathbb{F}_p] = [\mathbb{F}_{p^r} : \mathbb{F}_p[\alpha]][\mathbb{F}_{p}[\alpha] : \mathbb{F}_p] $](//latex.artofproblemsolving.com/1/0/3/1038bd4f769982d32144e5fce15bbe919696d985.png)
![$ \Longrightarrow r = [\mathbb{F}_{p^r} : \mathbb{F}_p[\alpha]]d $](//latex.artofproblemsolving.com/7/6/7/7672dad678f86089456bc8a3473731b4a212bcf2.png)
so 
as desired.
Now I shall show the "if" direction. If then 
is a subfield of Now let be a root of in the splitting field of over Then since ![$ [\mathbb{F}_p[\alpha] : \mathbb{F}_p] = d $](//latex.artofproblemsolving.com/c/9/e/c9ec8c48e18440bf4984c1c6c34833a968e5cc86.png)
we have ![$ \mathbb{F}_p[\alpha] = \mathbb{F}_{p^d}. $](//latex.artofproblemsolving.com/8/7/e/87e55e7a417f5e3c2845ff5ff6d73dca807e3ed9.png)
Hence 
which implies the desired result.
More to come later.
Well this immediately follows from 
and 
so let's solve those.
Let 
for some 
where 
for all 
Then 
in ![$ \mathbb{Z}_p[x] $](http://latex.artofproblemsolving.com/c/7/6/c760f9013d69b137936d845f8cd71f73d9f12974.png)
by the Frobenius endomorphism. And as it's clear that 
for all 
we obtain the desired result.
First I prove the "only if" direction. Let 
be a root of 
in the splitting field of over 
Note that since 
we have that 
and so ![$ \mathbb{F}_p[\alpha] $](http://latex.artofproblemsolving.com/9/7/4/97428c0562aa53c0e15c7c5bc93fa8d52f00b389.png)
is a subfield of 
Then by the Tower Law we find that ![$ [\mathbb{F}_{p^r} : \mathbb{F}_p] = [\mathbb{F}_{p^r} : \mathbb{F}_p[\alpha]][\mathbb{F}_{p}[\alpha] : \mathbb{F}_p] $](http://latex.artofproblemsolving.com/1/0/3/1038bd4f769982d32144e5fce15bbe919696d985.png)
![$ \Longrightarrow r = [\mathbb{F}_{p^r} : \mathbb{F}_p[\alpha]]d $](http://latex.artofproblemsolving.com/7/6/7/7672dad678f86089456bc8a3473731b4a212bcf2.png)
so 
as desired.Now I shall show the "if" direction. If
then 
is a subfield of Now let be a root of in the splitting field of over Then since ![$ [\mathbb{F}_p[\alpha] : \mathbb{F}_p] = d $](http://latex.artofproblemsolving.com/c/9/e/c9ec8c48e18440bf4984c1c6c34833a968e5cc86.png)
we have ![$ \mathbb{F}_p[\alpha] = \mathbb{F}_{p^d}. $](http://latex.artofproblemsolving.com/8/7/e/87e55e7a417f5e3c2845ff5ff6d73dca807e3ed9.png)
Hence 
which implies the desired result.More to come later.
, which implies that each root of 
of 
satisfies 
,
and then combine this with the fact that 
to reach the conclusion?
August 2016
June 2016