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Algebra III
Home Assignment 1
Subhadip Chowdhury
Problem 1
Consider the sequence of homomorphisms
n
fn : Z[[x]] → Z/p Z, fn
X
ai x
i
=
X
ai p i
mod pn
i<n
Clearly these maps are compatible with the canonical transition homomorphisms
ϕn : Z/pn+1 Z → Z/pn Z
hence defining the projective limit, we get that there is a unique homomorphism
f : Z[[x]] → lim Z/pn Z = Zp
←−
compatible with the fn ’s.
Now take any y ∈ Zp . Suppose the map Zp → lim Z/pn Z associates with y the sequence (yi )i∈N ∈ Z/pi Z.
←−
Let a0 = y1 ∈ Z. Next note that y2 ∈ Z/p2 Z and ϕ2 (y2 ) = y1 . So ∃a1 ∈ Z such that y2 = y1 + a1 p.
Similarly ϕn (yn ) = yn−1 implies ∃ai ∈ Z such that yn = yn−1 + an−1 pn−1 . Continuing similarly, we find
thatP∃ai ∀i ∈ N such that yn+1 = y1 + a1 p + a2 p2 P
+ . . . + an pn . Consider the
of Z[[x]] defined
P∞element
∞
∞
i
i
i
by i=0 ai x . Then clearly by construction, fn ( i=0 ai x ) = yn . So f ( i=0 ai x ) = y. Hence f is
surjective.
Next we want to prove that ker(f P
) is the principal P
ideal generated by x − p. So we have to show
that if we have a formal power series
ai xi such that i<n ai pi mod pn = 0 for all n ∈ N, then x − p
divides the formal power series.
For n = 1, we have a0 ≡ 0( mod p). Hence a0 = pα0 for some α0 ∈ Z.
Then for n = 2, we have a0 + a1 p ≡ 0( mod p2 ) ⇒ α0 + a1 ≡ 0( mod p). Hence there is some integer
α1 such that α0 + a1 = pα1 .
Continuing in similar way, we get that for any n ≥ 1, we have
pn αn−1 + an pn = a0 + a1 p + . . . + an pn ≡ 0(
mod pn+1 )
so that there is an integer αn with αn−1 + an = pαn . We can summarize these relations by
a0 = pα0 , an = pαn − αn−1 ∀n ≥ 1
Hence
X
a0 + a1 x + a2 x2 + . . . = (p − x)(α0 + α1 x + α2 x2 + . . .) ⇒ (x − p) ai xi
1
(?)
Algebra
Conversely if
isomorphism
Subhadip Chowdhury
P
ai xi ∈ (x − p), then (?) holds. Thus
P
i<n
Assignment 1
ai pi ≡ 0 mod pn . Thus we have a ring
∼
=
g : Z[[x]]/(x − p) → Zp
By construction g is clearly a continuous bijection. Now Z[[x]]/(x − p) = lim Z[xn ]/(pn , x − p), hence
←−
it is compact and Zp is Hausdorff since the p-adic valuation makes it a metric space. Thus g is a
continuous bijection from a compact set to a Hausdorff space, hence it is a homeomorphism. Hence g is
an isomorphism of topological rings.
Problem 2
Denote α by α1 . Thus we have a simple root α1 of P in Fp i.e. PP
(α1 ) ≡ 0 P
mod p. We want to show
i
that ∃α ∈ Zp such that P (α) = 0 in Zp and f1 (α) = α1 where fn ( ai p ) = i<n ai pi is the canonical
projection map from Zp to Z/pn Z.
For n ≥ 1, suppose we have an element αn ∈ Z/pn Z such that P (αn ) ≡ 0 mod pn and P 0 (αn ) 6≡ 0
mod p. We look for an element αn+1 = αn + tpn ∈ Z/pn+1 Z such that P (αn+1 ) ≡ 0 mod pn+1 . Note
that by expanding out the polynomial P we can write
P (αn + tpn ) = P (αn ) + tpn P 0 (αn ) + O(p2n )
Reducing both sides modulo pn+1 , we see that for P (αn+1 ) ≡ 0 mod pn+1 to hold, we need
0 ≡ P (αn + tpn ) = P (αn ) + tpn P 0 (αn )
mod (pn+1 )
since 2n ≥ n + 1. Now P (αn ) = zpn for some integer z. So,
0 ≡ (z + tP 0 (αn ))pn
mod pn+1 ⇒ 0 ≡ z + tP 0 (αn )
mod p
So putting
t=−
z
P 0 (α
n)
=−
P (αn )
n
p P 0 (αn )
we get an explicit formula:
αn+1 = αn −
P (αn )
P 0 (αn )
Since P 0 (αn ) is nonzero modulo p and hence pn , it has a unique inverse. Hence αn+1 is unique by
construction. Also note that by the first order Taylor expansion of P 0 at αn , we have
P 0 (αn+1 ) = P 0 (αn + (αn+1 − αn ))
= P 0 (αn ) + (αn+1 − αn ).(terms involving second and higher derivatives of P)
= P 0 (αn ) + (αn+1 − αn ).s (,let)
= u + tspn
where u is a unit in Z/pZ. Thus clearly P 0 (αn+1 ) 6≡ 0 mod p.
2
Algebra
Subhadip Chowdhury
Assignment 1
Thus by Induction principle, for all n ∈ N, we have constructed a series of element αn ∈ Z/pn Z
such that ϕn (αn ) = αn−1 where ϕn : Z/pn Z → Z/pn−1 Z is the canonical projection homomorphism and
P (αn ) ≡ 0 mod pn . Then taking projective limit we can find an element α ∈ Zp such that fn (α) = αn
for all n ∈ N and P (α) = 0 in Zp . Thus we have proved that P a zero α ∈ Zp of which, image in Fp is
α1 = α.
Problem 3
3.1
By the definition of topology on Af in , if
Q
pZp is open in Af in , then its intersection with n−1 Ẑ
p∈P
−1
must be open in n Ẑ for all n ∈ N. In particular for n = 1, we must have
Y
\
pZp Ẑ is open in Ẑ
p∈P
Now
Y
pZp
\
Ẑ =
p∈P
But the open subsets of
Q
Y
pZp
\Y
p∈P
p∈P
Zp =
Y
pZp
p∈P
Zp of the form
p∈P
Y
Ωp ×
Y
Zp
p6∈S
p∈S
Q
Q
Zp . Then any open subset of
Zp has
where S ⊆ P is a finite set; forms a basis of open subsets of
p∈P
p∈P
Q
pZp cannot be an open subset of Ẑ since pZp 6= Zp
infinitely many coordinates equal to Zp . Hence
p∈P
Q
and consequently,
pZp is not an open subset of Af in .
p∈P
Next we claim that Af in is a topological ring. Clearly addition is a continuous function from
Af in × Af in → Af in . To show that the multiplication operation Af in × Af in → Af in is continuous it is
enough to prove that multiplication by an element of Af in from n−1 Ẑ to m−1 Ẑ is a continuous map,
which is true.
Thus, in particular, multiplication by (p)p∈P ∈ Af in is a continuous function.
Note that
Q
Zp is compact by Tychonoff’s theorem and hence, the fact that Ẑ continuously injects into
Q
Af in implies it is a compact subset of Af in . Then the continuous image of
Zp under the multiplication
p∈P
Q
by the element (p)p∈P is also a compact set. Thus
pZp is a compact subset of Af in .
p∈P
p∈P
Another way of seeing this is to note that we can prove that the topology
Q on Af in is equivalent to the
product topology. Then by Tychonoff theorem, we can directly say that
pZp is a compact subset of
p∈P
Af in .
3
Algebra
Subhadip Chowdhury
Assignment 1
Note that Ẑ is product of Hausdorff spaces, and hence Hausdorff. Thus n1 Ẑ is Hausdorff for each n.
But given any two points Q
in Af in , we can find N ∈ N such that they are elements of N −1 Ẑ. Thus Af in is
a Hausdorff space. Then
pZp is a compact subset of a Hausdorff space and hence a closed subset
p∈P
of Af in .
Q
Another way of seeing this is to note that the complement of
pZp in Ẑ is
p∈P
[
p∈P
which is open and hence
Q
1
z ∈ Zp : |z|p >
p
!
×
Y
Zq
p6=q∈P
pZp is a closed subset of Ẑ. But Ẑ is a closed subset of Af in . Hence
p∈P
Q
pZp
p∈P
is a closed subset of Af in .
3.2
S
Note that Ẑ is an open subset of Af in . It is also closed since its complement
n−1 Ẑ is open.
n∈N
n>1
Now the given set Af in ∩
−1
Q
Q
p Zp is the inverse image of
Zp under the multiplication map by
p∈P
Q −1
∈ Af in , which is continuous. Hence Af in ∩
p Zp is an open and closed subset of Af in
p∈P
(p)p∈P
p∈P
Suppose Af in ∩
Q
p−1 Zp is compact subset of Af in . Note that n−1 Ẑ is open in Af in for each n ∈ N,
p∈P
by definition of the topology on Af in . Now clearly, the collection of these open sets
[
n−1 Ẑ
n∈N
n−1 Ẑ. Then by compactness, there is a finite
Q −1
subcollection from above collection of open sets which covers Af in ∩
p Zp . Suppose
forms an open cover of Af in ∩
p−1 Zp ; since Af in =
Q
p∈P
S
n∈N
p∈P
Af in ∩
Y
−1
p Zp ⊆
k
[
n−1
i Ẑ
i=1
p∈P
But clearly above result cannot be true. Since for any p >> max{n1 , . . . , nk } we have
(1, 1, 1, . . . , 1,
Y
1
1
p−1 Zp but (1, 1, 1, . . . , 1,
, 1, . . .) ∈ Af in ∩
, 1, . . .) 6∈ n−1
i Ẑ
p
p
p∈P
|{z}
|{z}
pth term
Contradiction!! Hence Af in ∩
pth term
Q
p−1 Zp is not a compact subset of Af in .
p∈P
4