Today was the first day after my prelims that I met Dr. Shahidi. He earlier advised me to take a break after the advanced topics so I could start research afresh. I also recently found his Wikipedia page (albeit a stub) – http://en.wikipedia.org/wiki/Freydoon_Shahidi . A more relevant Wiki-link about his work is – http://en.wikipedia.org/wiki/Langlands%E2%80%93Shahidi_method I hope one day I’ll edit these pages to add sources and more relevant material.
Our conversation was quite pleasant. He had some things in mind he wanted me to work on. He told me to go over “L-packets” and “A-packets” and other technical stuff; gave references where I could read about it. We discussed about my general PhD goal. “Your (research) problem should not be too difficult to be unable to solve”, he said. On asking if I’d like my work to be “Algebra or Analysis”, I instantly replied “Algebra!” although I should know that Number Theory uses tools from all branches of mathematics (including PDE 
)
I have known Shahidi for his dry sarcastic wit and today’s conversation ended with a remarkable quip. He told me, “Keep me informed, don’t run away!”
. The problem boiled down to determining how a prime split in an extension, which reduced to finding the ring of integers of a number field. The usual method (following Kummer) is to find an algebraic integer that generates the ring of integers of the extension over the ring of integers of the base field. But such an integer is not guaranteed to exist. I posted this on math.stackexchange and got an interesting answer in the form of a comment which I now blog.
and 
be linearly disjoint number fields, i.e.,
, where 
denotes the discriminant of 
, etc. and ![{r = [K:\mathbb Q]}](http://s0.wp.com/latex.php?latex=%7Br+%3D+%5BK%3A%5Cmathbb+Q%5D%7D&bg=ffffff&fg=000000&s=0 "{r = [K:\mathbb Q]}")
and ![{s = [L : \mathbb Q]}](http://s0.wp.com/latex.php?latex=%7Bs+%3D+%5BL+%3A+%5Cmathbb+Q%5D%7D&bg=ffffff&fg=000000&s=0 "{s = [L : \mathbb Q]}")
. Moreover, if 
are relatively prime over 
, then 
and 
is an isomorphism. So, the basis of 
can be obtained from the bases of 
and 
in a natural way.
be the image of
is the image of a pure tensor 
, then the trace form on 
acts on 
by 
. (For a proof of this, go to the Galois closure and use the canonical isomorphism 
given by linear disjointness). Thus, 
acts by 
. The matrix of 
is then the Kronecker product of the matrices of 
and 
. It follows that 
. By transitivity of discriminants (see below), 
and 
, we also know that 
is divisible by 
and 
, hence the assertion that 
.
be a finite extension of number fields and 
be their rings of integers. Then the discriminants satisfy![\displaystyle D_{A''/A} = D_{A'/A}^{[K'':K']}.\mathrm{N}_{ K'/ K}(D_{ A''/ A'}).](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+D_%7BA%27%27%2FA%7D+%3D+D_%7BA%27%2FA%7D%5E%7B%5BK%27%27%3AK%27%5D%7D.%5Cmathrm%7BN%7D_%7B+K%27%2F+K%7D%28D_%7B+A%27%27%2F+A%27%7D%29.+&bg=ffffff&fg=000000&s=0 "\displaystyle D_{A''/A} = D_{A'/A}^{[K'':K']}.\mathrm{N}_{ K'/ K}(D_{ A''/ A'}).")
, the group of 
invertible matrices over a field 
, or any reductive connected closed subgroup of this matrix group.
, a Tits system (named after 
subgroups of 
pair
is a set such that the following axioms hold:
,
(called the 
for all 
,
and 
,
,
generate 
is a group.
as 
, where $w=nT$ and observe the well-definedness.
.
.
be a Tits system and 
be the Weyl group. Then,
.
. Prove that $G’$ is a group, and in fact a subgroup of 
contains 
and 
which generate 
.
whenever 
.
, which is a torus. One does however need to show that this forms a Tits system.
means. Intuitively, it means there is a knot at 
. 
. Let 
be an element of the function field 
that generates the maximal ideal of the local ring (also a dvr) ![{k[C_2]_Q}](http://s0.wp.com/latex.php?latex=%7Bk%5BC_2%5D_Q%7D&bg=ffffff&fg=000000&s=0 "{k[C_2]_Q}")
. Let 
be the 
as a function of 
. Then 
is ramified at 
and unramified if 
for every point 
.
be a non-constant map of curves. Then,
,
denotes the degree of separability of 
.
is another non-constant map of curves, then
![{\displaystyle\sum_{i=1}^g e_i f_i = [L:K]}](http://s0.wp.com/latex.php?latex=%7B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eg+e_i+f_i+%3D+%5BL%3AK%5D%7D&bg=ffffff&fg=000000&s=0 "{\displaystyle\sum_{i=1}^g e_i f_i = [L:K]}")
for number fields 
. The second one says that only finitely many primes ramify and the third result is the multiplicativity of ramification indices in a tower of number fields. Let us state these results more precisely in the following
be number fields with ![{[L:K]< \infty}](http://s0.wp.com/latex.php?latex=%7B%5BL%3AK%5D%3C+%5Cinfty%7D&bg=ffffff&fg=000000&s=0 "{[L:K]< \infty}")
. Let 
‘s are primes in ![\displaystyle \displaystyle\sum_{i=1}^g e_i f_i = [L:K]](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eg+e_i+f_i+%3D+%5BL%3AK%5D+&bg=ffffff&fg=000000&s=0 "\displaystyle \displaystyle\sum_{i=1}^g e_i f_i = [L:K]")
holds, where 
is the inertial degree given by ![{[ \mathcal O_L/Q_i : \mathcal O_K/P ]}](http://s0.wp.com/latex.php?latex=%7B%5B+%5Cmathcal+O_L%2FQ_i+%3A+%5Cmathcal+O_K%2FP+%5D%7D&bg=ffffff&fg=000000&s=0 "{[ \mathcal O_L/Q_i : \mathcal O_K/P ]}")
.
for some 
).
is the group of units)
to be algebraically closed. (This is far from sufficient; indeed the most interesting applications in number theory will have 
-adic fields). One can think of the curve as the locus of the zero set of an irreducible polynomial ![{F(X,Y) \in k[X,Y]}](http://s0.wp.com/latex.php?latex=%7BF%28X%2CY%29+%5Cin+k%5BX%2CY%5D%7D&bg=ffffff&fg=000000&s=0 "{F(X,Y) \in k[X,Y]}")
. This as it is, is an affine curve and we attach points of infinity as necessary by homogenizing 
. Note that it is not necessary that the curve be generated by just one polynomial. It may be possible that it is the intersection of higher dimensional varieties in projective space of higher dimension. The only requirement is that the dimension of this variety be 1.
, we can associate a field to it, namely the function field of 
. (If you are not habituated to using projective coordinates, just substitute 1 for 
and everything should work fine). Then the function field of ![\displaystyle k(C) = \text{Field of fractions of } \displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}.](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+k%28C%29+%3D+%5Ctext%7BField+of+fractions+of+%7D+%5Cdisplaystyle%5Cfrac%7Bk%5BX%2CY%2CZ%5D%7D%7B%28X%5E2%2BY%5E2-Z%5E2%29%7D.+&bg=ffffff&fg=000000&s=0 "\displaystyle k(C) = \text{Field of fractions of } \displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}.")
having a transcendence degree 1 over ![{P=[a:b:c]}](http://s0.wp.com/latex.php?latex=%7BP%3D%5Ba%3Ab%3Ac%5D%7D&bg=ffffff&fg=000000&s=0 "{P=[a:b:c]}")
on the curve, there is a discrete valuation ring ![{k[C]_P}](http://s0.wp.com/latex.php?latex=%7Bk%5BC%5D_P%7D&bg=ffffff&fg=000000&s=0 "{k[C]_P}")
which is the localization of the domain ![{\displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}}](http://s0.wp.com/latex.php?latex=%7B%5Cdisplaystyle%5Cfrac%7Bk%5BX%2CY%2CZ%5D%7D%7B%28X%5E2%2BY%5E2-Z%5E2%29%7D%7D&bg=ffffff&fg=000000&s=0 "{\displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}}")
at the maximal ideal 
. A fundamental theorem in algebraic geometry says that the point 
and 
we will mean a morphism of the corresponding projective varieties. If
![{[k(C_1) : \phi^*(k(C_2))]}](http://s0.wp.com/latex.php?latex=%7B%5Bk%28C_1%29+%3A+%5Cphi%5E%2A%28k%28C_2%29%29%5D%7D&bg=ffffff&fg=000000&s=0 "{[k(C_1) : \phi^*(k(C_2))]}")
and we define the degree 
to be the degree of this extension. A map of degree 1 is an isomorphism.
of transcendence degree 1 over 
of transcendence degree 1, one proves that the collection of local rings 
such that 
actually define a non-singular projective curve. (All the 
). The morphisms in this category are inclusion maps of fields and they give morphisms of curves.
miles
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