Report Follows Very Closely the Book of Svetlana Katok1

Elementary Analysis in Qp

Hannah Hutter, May Szedl´ak,Philipp Wirth November 17, 2011

This report follows very closely the book of Svetlana Katok1.

In this section we will see some properties of sequences and series in Qp. Recall from the ﬁrst presentation that Qp is a complete metric space. Hence every Cauchy sequence converges and therefore the set of the convergent sequences is the set of the Cauchy sequences. We will also see that we have some much better properties in Qp, than we are used to in the real case. The ﬁrst example is the characterization of the Cauchy sequences.

Theorem 1. A sequence {an} in Qp is a Cauchy sequence (i.e. converges), if and only if lim |an+1 − an|p = 0. n→∞ Pn 1 Remark. This is obviously not true in R. For example take an = k=1 k 1 the n-th harmonic number. Then limn→∞ |an+1 −an|p = limn→∞ n+1 = 0 but as we know the an is not Cauchy.

Proof. Remember that if {an} is a Cauchy sequence, then

lim |am − an|p = 0. m,n→∞

The ﬁrst direction follows immediately for m = n + 1. Note that this is true for Cauchy sequences in any metric space.

1Katok, Svetlana; p-adic Analysis Compared with Real, American Mathematical Soci- ety, 2007

1 For the converse assume limn→∞ |an+1 − an|p = 0. This means that for any > 0 there exists N ∈ N s.t. for n > N

|an+1 − an|p < . Then for any m > n > N, using the strong triangle inequality, we get

|am − an|p = |am − am−1 + am−1 − am−2 + · · · − an|p

≤ max(|am − am−1|p,..., |an+1 − an|p) < , which completes the proof. P∞ Deﬁnition 1. Let the series n=1 an be in Qp. The sum converges if the sequence of its partial sums converges in Qp, i.e. PN if limN→∞ |SN+1 − SN |p = 0 where SN = n=1 an. P∞ The sum converges absolutely if n=1 |an|p converges in R. Example. We have ∞ X n2 · (n + 1)! = 2 n=1 in Qp, for any p. By induction we will show that for the N partial sum one has

N X n2 · (n + 1)! = 2 + (N − 1) · (N + 2)!. n=1 Then ∞ X 2 n · (n + 1)! − 2 = lim |2 + (N − 1) · (N + 2)! − 2|p = 0 p N→∞ n=1 since |(N + 2)!|p → 0 for N → ∞. The proof by induction is simple calculus: Case N = 1: 1 X n2 · (n + 1)! = 1 · 2 = 2 = 2 + (1 − 1) · (1 + 2)!, n=1 hence the induction base holds. For N → N + 1: N+1 N X X n2 · (n + 1)! = n2 · (n + 1)! + (N + 1)2 · (N + 2)! n=1 n=1 =2 + (N − 1) · (N + 2)! + (N + 1)2 · (N + 2)! =2 + (N + 2)! · (N − 1 + N 2 + 2N + 1) =2 + N · (N + 3)!.

2 As we are used to in R, convergence follows from absolute convergence by the triangle inequality. P∞ Proposition 1. If the series i=1 |ai|p converges absolutely (in R), then P∞ i=1 ai converges in Qp. P∞ Proof. Suppose that i=1 |ai|p converges. Then the sequence of its partial sums is Cauchy i.e. for any > 0 there exists N ∈ N s.t. for all m > n > N, Pm we have i=n+1 |ai|p < . By the usual triangle inequality we get

m m X X |Sm − Sn|p = ai ≤ |ai|p < . p i=n+1 i=n+1 P∞ Therefore {Sn} is a Cauchy sequence and by deﬁnition i=1 ai converges in Qp.

We however can get a much better result in Qp as seen in the next proposition. P∞ Proposition 2. A series i=1 ai in Qp converges in Qp if and only if limi→∞ ai = 0 and in this case

∞ X ai ≤ max |ai|p. p n i=1 Remark. The last property can be considered as the strongest wins property for converging series. Remark. We again know that this proposition is not true in R. By taking 1 ai = i , we can see that limi→∞ ai = 0 but again the corresponding series does not converge. P∞ Proof. We know by deﬁnition that i=1 ai converges if and only if the se- P∞ quence of partial sums Sn = i=1 ai converges. Since an = Sn − Sn−1 the series converges if and only if an goes to 0. For the second part we assume P∞ P∞ that i=1 ai converges. In the case where i=1 ai = 0 the proof is trivial. P∞ Now suppose i=1 ai 6= 0. Then by the strong triangle inequality for any partial sum we have n X ai ≤ max |ai|p. p 1≤i≤n i=1

Since ai → 0 for n → ∞ for large enough N we have

max |ai|p = max |an|p. 1≤i≤N n

3 Pn P∞ Since the partial sum i=1 ai converges to i=1 ai 6= 0 and the p-adic norm can only take a discrete amount of values, for m large enough we have

Pm P∞ i=1 ai = i=1 ai . The claim follows by taking the maximum of m p p and n. P∞ Deﬁnition 2. A series n=0 an converges unconditionally if for any re- 0 P∞ 0 ordering of the terms an → an the series n=0 an also converges. P∞ Theorem 2. If n=0 an converges, it converges unconditionally and the sum does not depend on the reordering.

Remark. This is not true in R. We can even prove that for any convergent P∞ but not absolutely convergent series n=0 an we have that for every x ∈ R 0 P∞ 0 there exists a reordering of the summands an → an such that n=0 an = x.

Proof. Let > 0 and let N ∈ N s.t. for any n > N we have that |an|p < , 0 |an| < and ∞ N X X an − an < . p n=1 n=1

We know that we can find such an N since by Proposition 2 the terms an PN 0 PN 0 tend to zero. Let S = n=1 an and S = n=1 an and define by S1 and 0 S1, respectively, the sum of all terms in S for which |an|p ≥ and the sum 0 0 of all terms of S for which |an|p ≥ . By construction it is clear that S1 0 0 and S1 contain the same terms and therefore S1 = S1. The sum S differs 0 0 from S1 only by the terms an satisfying |an|p < and similarly S from S1 0 0 by the terms an satisfying |an|p < . Hence by applying the strong triangle 0 0 0 inequality we get |S − S1|p < and |S − S1|p < and therefore |S − S |p < . Hence we have

∞ N ∞ N N N X X 0 X X X X 0 an − an = an − an + an − an < . p p n=1 n=1 n=1 n=1 n=1 n=1

P∞ 0 Now for → 0 we see that the series n=1 an converges and

∞ ∞ X X 0 an = an n=1 n=1 as desired.

4 Remark. As in R, convergence does not imply absolute convergence. Con- sider the following series: 1; p repeated p times; p2 repeated p2 times; . . . . Since the terms converge to 0 the series converges but

∞ X −1 2 −2 |an|p = 1 + p · p + p · p + ··· = ∞, i=1 and hence the series does not converge absolutely. Conclusion. Summing up the previous theorems and remarks we can conclude the following properties. In Qp one has the follwing implications: absolute convergence ⇒ convergence ⇔ unconditional convergence. On the other hand in R one has: absolute convergence ⇒ unconditional convergence ⇒ convergence. Note that diﬀerent from Qp in R: convergence ; unconditional convergence. In both cases one has: convergence, unconditional convergence ; absolute convergence.

The next proposition is pretty subtle in the real case and stated here because it will be used later.

Theorem 3. Let bij ∈ Qp, i, j = 1, 2,... , such that for any > 0 there exists N ∈ N such that

max(i, j) ≥ N ⇒ |bij|p < i.e. |bij|p tends to zero if either i or j goes to inﬁnity. Then the two series

∞ ∞ ∞ ∞ X X X X bij and bij i=1 j=1 j=1 i=1 converge, and their sums are equal. P∞ Proof. We know from Proposition 2 that j=1 bij converges for all i and P∞ i=1 bij converges for all j. By the same proposition we know that for all i ≥ N ∞ X bij ≤ max |bij|p < p j j=1 and for all j ≥ N ∞ X bij ≤ max |bij|p < . p i i=1

5 This implies that both double series converge, again by Proposition 2. It is therefore left to show that both sums are equal. We know that

∞ ∞ N N X X X X bij − bij p i=1 j=1 i=1 j=1 N ∞ ∞ ∞ X X X X = bij + bij p i=1 j=N+1 i=N+1 j=1 N ∞ ∞ ∞ n X X X X o ≤ max bij , bij < , p p i=1 j=N+1 i=N+1 j=1 where the last inequality follows from Proposition 2. Since this is true for any , the series must be equal.

2 p-adic Power Series

Definition 3. As in the real case a formal power series is of the form ∞ X n f(X) = anX , n=0 where the coefficients an are in Qp and X is the indeterminate. The set of all power series in X with coefficients in Qp are denoted by Qp[[X]] and the set of all polynomials by Qp[X].

Let x ∈ Qp and f ∈ Qp[[·]]. Then the corresponding power series f(x) is P∞ n n=0 anx . From Proposition 2 we know that f(x) converges if and only if n n |anx |p → 0. Therefore if for a 0 ≤ r ∈ R we have that |an|pr → 0, our P∞ n series n=0 anx converges for any x ∈ Qp s.t. |x|p ≤ r. P∞ n Deﬁnition 4. For a power series f(X) = n=0 anX with an ∈ Qp the radius of convergence is deﬁned as the extended real number 0 ≤ rf ≤ ∞ s.t.

n rf = sup{r ≥ 0 : |an|pr → 0}.

The extended real numbers are deﬁned as R ∪ {∞}. The convergence radius can be computed as in the Archimedian case. P∞ n Proposition 3. For a power series f(X) = n=0 anX the radius of convergence is 1 rf = 1 . n lim supn |an|p

6 Proof. The proof works as in the real case. Suppose that |x|p > rf (in the case where rf < ∞). We have

1 1 k k 1 lim sup |x|p|ak|p = |x|p lim sup |ak|p = |x|p · > 1. rf k k Therefore for inﬁnitely many values k we have that |ak|p|x|p > 0 i.e. akx does not converge to 0 and hence the series diverges. Now suppose that |x|p < rf (in the case where rf > 0). Then we can choose r ∈ R such that |x|p < r < rf . Then

1 1 k k r lim sup r|ak|p = r lim sup |ak|p = < 1. rf

Hence there exists an N ∈ N such that

1 k sup r|ak|p < 1. k≥N

k Therefore for k > N we have |ak|pr < 1 and

|x| k |x|k |a xk| = |a | rk p < p → 0 for k → ∞. k p k p r rk

Since the terms of the series go to 0 it follows that the series converges. The natural question that arises is, whether the series converges on the boundary or not.

Proposition 4. The domain of convergence of a p-adic power series f(X) ∈ k Qp[[X]] is a ball D = {|x|p ≤ R} where R ∈ {p , k ∈ Z} ∪ {0} ∪ {∞}, and the series converges uniformly in D.

Remark. We know that in the Archemedian case we can not make a statement about the convergence on the boundary. Take for example the power P∞ n+1 n series of ln. We know that ln(1 + x) = n=1(−1) x /n. One can ﬁnd the the radius of convergence is 1, but for x = −1 the series diverges and for x = 1 the series converges.

Proof. Suppose the radius of convergence of f(X) is rf . As seen before the series converges on the open ball D = {|x|p < R}. We know that on the n boundary i.e. where |x|p = rf , f(X) converges if and only if |anx |p → 0. We can observe that this depends only on the norm of x and not the speciﬁc value of x, hence either for all points with |x|p = rf the series converges or it does for none. If the series converges for all points with |x|p = rf then we

7 have R = rf . If the series diverges for all points with |x|p = rf , then we have −1 R = p rf . Finally for any |x|p ≤ R we have

n n |anx |p ≤ |anR |p → 0 and therefore the uniform convergence on D follows.

3 p-adic Logarithm and Exponential

Deﬁnition 5. Similar as in the real setting, we deﬁne the p-adic logarithm by the series ∞ n X n+1 (x − 1) ln (x) := (−1) . p n n=1 n o This series converges for all x ∈ B := x ∈ Zp| |x − 1|p < 1 = 1 + pZp. Note that the p-adic logarithm corresponds to the formal power series

∞ n X n+1 X log (1 + X) = (−1) ∈ [[X]] . n Q n=1 Theorem 4. The p-adic logarithm satisﬁes the fundamental property

lnp (xy) = lnp (x) + lnp (y) for all x, y ∈ B such that xy is in B as well. Proof. Consider the following identity of formal power series:

log (1 + X) + log (1 + Y ) − log (1 + X + Y + XY ) = 0.

Clearly, this holds for the real logarithmic function, i.e. the left hand side of the above identity vanishes for all real numbers x, y ∈ (−1, 1). We also know that all the series converge absolutely, which means that we can rearrange P i j the terms such that the left side is of the form ci,jx y with ci,j = 0 for all i, j ∈ N. Now we choose any α, β ∈ pZp. Clearly, α + β + αβ ∈ pZp and by applying Theorem 3, the expression

lnp (1 + α) + lnp (1 + β) − lnp ((1 + α) (1 + β))

P i j can be written in the form ci,jα β where all ci,j = 0. This implies the theorem.

8 Deﬁnition 6. Consider the formal power series

∞ X Xn exp (X) = n! n=0 the corresponding power series in Qp is called the p-adic exponential and is denoted by expp (x). Unfortunately, the radius of convergence of the p-adic exponential is not as obvious as the one of the p-adic logarithm. To ﬁnd this, we need the following

Lemma 1. For any positive integer n ∈ N, let

k X i n = aip i=0 be its expansion in base p. We then have n − S ord (n!) = n , p p − 1 where Sn = a0 + a1 + ... + ak.

n−Sn Proof. If p > n, we have Sn = n and therefore p−1 = 0. Of course this is true since p - n! and therefore ordp (n!) = 0. Now let us assume p ≤ n and Pk i let n = i=0 aip be the expansion of n in base p. For any 1 ≤ i ≤ k there are exactly j n k j n k − pi pi+1 numbers between 1 and n such that pi is the is the greatest power of p dividing each. So we get

k k X j n k j n k X j n k ord (n!) = i − = . p pi pi+1 pi i=1 i=1 Here the last equation holds because for i ≥ 2, each term appears twice in the sum, once with coefficient i and once with coefficient −(i + 1). Further, j n k pk+1 = 0 by the definition of k.

9 On the other hand we have n − S a + a p + ... + a pk − (a + a + ... + a ) n = 0 1 k 0 1 k p − 1 p − 1 a (p − 1) + a (p2 − 1) + ... + a pk − 1 = 1 2 k p − 1 k k−1 ! X X j X j = ai p = aip i=1 j=0 1≤j

+ ak k jnk j n k j n k X j n k = + + = p p2 pk pi i=1 as desired.

Theorem 5. The p-adic exponential expp converges in the disc

n −1/(p−1)o Dp := x ∈ Qp| |x|p < p .

P n Proof. The convergence radius rp of a power series anX is given by the formula 1 rp = 1/n . lim sup |an|p

In our case, an = 1/n!. Thus we get by Lemma 1

1 n − S 1 ord (r ) = − lim inf ord (a ) = lim inf n = , p p n p n n (p − 1) p − 1 where Sn is the same as in Lemma 1 and the last equality holds since n − S S lim − n = −1 + lim n = −1. n→∞ n n→∞ n

Since rp is a power of p, we have

ordp(rp) −1/(p−1) rp = p = p .

10 −1/(p−1) Remark. If |x|p = p then we have xn n − S n S ord = − n + = n . p n! p − 1 p − 1 p − 1

k nk Now consider the subsequence nk = p . Then Snk = 1 and ordp (x /nk!) = 1 p−1 . Hence we have

nk x −1/(p−1) lim 6= 0 for |x|p = p . k→∞ nk! p

This means that expp (x) diverges on ∂Dp. Proposition 5. The p-adic exponential satisﬁes the fundamental property

expp (x + y) = expp (x) expp (y) for all x, y ∈ Dp. Proof. Similar as in Theorem 4, the result is true for formal power series. Therefore we can conclude the result exactly the same way as in Theorem 4.

21 22 23 ln (−1) = ln (1 − 2) = − + + + ... . 2 2 1 2 3 On the other hand, we know from the fundamental property of the p-adic logarithm that

0 = ln2 (1) = ln2 (−1) + ln2 (−1) = 2 ∗ ln2 (−1) .

22 23 This means that the sum 2 + 2 + 3 + ... converges to 0 as n → ∞. The next goal is to show that the p-adic logarithm and the p-adic exponential n o are inverse to each other in some sense. Again, let Dp = x ∈ Zp| |x|p < rp −1/(p−1) be the region of convergence of expp, where rp = p . Before we can prove the main result, we need the following estimates:

Lemma 2. Let n ≥ 2 and let 0 < |x|p < rp. Then the following estimate holds: n n x x ≤ < |x|p < rp. n p n! p

11 Proof. From Lemma 1 we know that for any integer n ≥ 1, n − S n − 1 ord (n!) = n ≤ . p p − 1 p − 1

But since ordp (n) ≤ ordp (n!), we clearly have

n−1 − p−1 n−1 |n|p ≥ |n!|p ≥ p = rp . Therefore, by combining these two estimates, we get

n n n−1 x x |x|p ≤ ≤ |x|p < |x|p < rp. n p n! p rp

Proposition 6. The maps

expp : Dp → 1 + Dp and lnp : 1 + Dp → Dp are inverse to each other. This means that for all x ∈ Dp,

lnp (expp (x)) = x and expp (lnp (1 + x)) = 1 + x.

Proof. By considering the corresponding formal power series, we know that the relations in the Proposition hold. So the only thing we have to make sure is that all involved power series converge. In other words, we have to show that expp (x) ∈ 1 + Dp and lnp (1 + x) ∈ Dp for any x ∈ Dp. Then the convergence follows from the discussion above. By applying the estimate form the previous Lemma and the strongest wins property for series, we have

∞ X xn |exp (x) − 1| = x + ≤ |x| < r . p p n! p p n=2 p

The last equality holds because all the terms in the sum are less than |x|p by Lemma 2. Therefore, expp (x) ∈ 1 + Dp and expp : Dp → 1 + Dp. The same argument works for the p-adic logarithm. Let 1 + x ∈ 1 + Dp. Then ∞ n X n+1 x |ln (1 + x)| = x + (−1) ≤ |x| < r . p p n p p n=2 p Again, the last equality follows from Lemma 2 and the strongest wins property for series. Thus we have lnp (1 + x) ∈ Dp for all x ∈ Dp and therefore lnp : 1 + Dp → Dp.

12 n o Now let x, y ∈ Dp and let z ∈ Zp. Then |x + y|p ≤ max |x|p , |y|p and

|zx|p ≤ |x|p. Therefore, x + y and zx are in Dp as well, which means that Dp is an ideal in Zp. In particular, Dp is an additive subgroup of Zp. Further, x + y + xy ∈ Dp, which means that (1 + x) (1 + y) = 1 + x + y + xy ∈ 1 + Dp. This means that 1 + Dp is a multiplicative subgroup of Zp. Proposition 7. The p-adic logarithm is an isomorphism of groups

expp : Dp → 1 + Dp.

Proof. From the last Proposition, we already know that expp is bijective. Therefore it is enough to show that expp is a group homomorphism. But this follows directly from the fundamental property:

expp (x + y) = expp (x) expp (y) .

Until now, the p-adic logarithm and exponential behave quiet similar as the corresponding maps in the setting of real numbers. But there is one property which is totally diﬀerent in the p-adic setting. In fact, the p-adic exponential is an isometry! Recall that an isometry is a map that preserves the distance. To prove this, we need the following Proposition:

Proposition 8. Let x ∈ Dp. Then the following holds:

1. |expp (x)|p = 1,

2. |lnp (1 + x)|p = |x|p,

3. |1 − expp (x)|p = |x|p.

Proof. Let x ∈ Dp. By the estimate of Lemma 2, the term of maximal norm in the series ∞ X xn exp (x) = 1 + x + p n! n=2 is 1. By the strongest wins property for series we get that |expp (x)|p = 1. Exactly the same way we see that in the series

∞ X xn exp (x) − 1 = x + , p n! n=2 the term of maximal norm is x and therefore |1 − expp (x)|p = |x|p.

13 Similarly, the term of maximal norm in the series

∞ n X n+1 x ln (1 + x) = x + (−1) p n n=2 is x and thus |lnp (1 + x)|p = |x|p. Corollary 1. The maps

expp : Dp → 1 + Dp and lnp : 1 + Dp → Dp are isometries.

Proof. Let x, y ∈ Dp. Then we have

|expp (x) − expp (y)|p = |expp (y)|p |expp (x − y) − 1|p

= |expp (x − y) − 1|p = |x − y|p , where the second equation follows from part 1 and the last equation follows from part 3 of the previous proposition. This shows that expp is an isometry. To show the same for the p-adic logarithm, we use that expp (lnp (1 + x)) = 1 + x for x ∈ Dp. Then, since

|lnp (1 + x) − lnp (1 + y)|p = |expp (lnp (1 + x)) − expp (lnp (1 + y))|p

= |(1 + x) − (1 + y)|p = |x − y|p , the p-adic logarithm is an isometry.

4 Strassman’s Theorem

This last section is about the number of zeros of functions given by Zp- converging power series. First the Strassman’s theorem is stated, which gives us an upper bound for the number of zeros for certain functions. Then some interesting corollaries will follow.

Theorem 6 (Strassman’s theorem). Let f : Zp → Qp, x 7→ f(x), be given P n by a nonzero power series f(X) = n≥0 anX ∈ Qp[[X]]. Suppose that limn→∞ an = 0, so that f(x) converges for all x ∈ Zp. Now let N ∈ N0 be deﬁned by:

1. |aN |p = maxn≥0 |an|p, (since limn→∞ an = 0 |an|p attains its maximum for a ﬁnite set of indices).

14 2. |an|p < |aN |p ∀n > N, i.e., N is the largest index of an with maximum norm.

Now one can conclude that f has at most N zeros. Before proving Strassman’s Theorem an example: P n Example. Consider the power series f(X) = n≥0 n!X . We want (1) to determine the radius of convergence and (2) to estimate the number of zeros from above. −1 (1) Set an := n!. |an|p = 1 for all n ∈ {0, . . . , p − 1}, |an|p ≤ p for all n ∈ {p, . . . , 2p−1} and so on. So lim supn≥0 |an|p = 1, hence the convergence n n→∞ radius is 1. Since |an1 |p −→ 0, f converges in Zp. (2) The number we are looking for is N = p−1. We are then able to conclude by applying Stassman that f has at most p − 1 zeros. Remark. In R this theorem is clearly not true. Consider to see this X (−1)n f(X) = sin(X) = X2n+1. (2n + 1)! n≥0

The coeﬃcients an are given by

(0, if n even, a = n−1 n (−1) 2 n! , else. From this, one can see that N = 1. But we know that sine has inﬁnitely many zeros and not at most one! Proof. The proof is done by induction on N. N = 0: We know from the condition on N that |an|p > |a0|p for all n > 0. Now assume for the purpose of deriving a contradiction that there exists x ∈ Zp such that

X n 2 0 = f(x) = anx = a0 + a1x + a2x .... n≥0

P n We obtain that |a0|p = | n≥1 anx |p, which can be rewritten as the following since we can use the ’strongest wins’-property for sums:

X n n |a0|p = | anx |p ≤ max |anx |p. n≥1 n≥1

Finally we get (since |x|p ≤ 1):

n |a0|p ≤ max |anx |p ≤ max |an|p < |a0|p, n≥1 n≥1

15 which is a contradiction. Our assumption was proved wrong so there exists no x ∈ Zp such that f(x) = 0. This shows the base of induction. N ≥ 1: Assume the statement of the theorem holds for m < N. We show that it also holds for N. To do this we factor out one zero and show that we are allowed to use the induction hypothesis on our new power series. Assume that f has at least one zero in Zp denoted by α. Otherwise the proof is trivial (0 < N). Suppose N is given as in the assumption of the theorem. Choose x ∈ Zp arbitrary. We get

n−1 X n n X X j n−1−j f(x) = f(x) − f(α) = an(x − α ) = (x − α) anx α , n≥0 n≥0 j=0 where the last equality can be justiﬁed by multiplying out

n−1 X (x − α) xjαn−1−j = xn − αn. j=0

Since x and α are elements of Zp, we have |x|p, |α|p ≤ 1. Hence we obtain j n−1−j n→∞ that |anx α |p ≤ |an|p −→ 0. We can use Theorem 3(interchanging sums) and achieve that

n−1 ∞ X X j n−1−j X X j n−1−j f(x) = (x − α) anx α = (x − α) anx α n≥0 j=0 j≥0 n=j+1

= (x − α)g1(x), where ∞ ∞ X j X n−1−j X k g1(x) = bjx , bj = anα = aj+1+kα . j≥0 n=j+1 k=0

Now we show that (1) g1 converges in Zp and (2) the largest index of bj with maximal norm is N − 1. j→∞ (1) g1 converges in Zp if |bj|p −→ 0 (theorem from before). This is indeed the case (again with the ’strongest wins’-property):

∞ X k k |bj|p = | aj+1+kα |p ≤ max |aj+1+kα |p k≥0 k=0 j→∞ ≤ max |aj+1+k|p −→ 0. k≥0

(2) We have |bj|p ≤ max |aj+1+k|p ≤ |aN |p ∀j ∈ N0, k≥0

16 To obtain |bN−1|p = |aN |p one has to work a bit more. We have:

n X k X k |bN−1|p = | aN+kα |p = lim | aN+kα |p, n→∞ k≥0 k=0 where the last equality holds because of continuity of the norms. By the ’strongest-wins’-property from the ﬁrst talk, i.e., if |a|p < |b|p then |a + b|p = k |b|p and since |aN+kα |p ≤ |aN+k|p < |aN |p ∀k ≥ 1, we have that

n X k |bN−1|p = lim | aN+kα |p = |aN |p. n→∞ k=0

∀j ≥ N : |bj|p ≤ max |aj+1+k|p = max |an|p < |aN |p. k≥0 n≥N+1

So bN−1 has maximal norm and N − 1 is the largest index with maximal norm. We are now allowed to use the induction hypothesis on g1, hence g1 has at most N − 1 zeros. We conclude that f has at most N − 1 + 1 = N zeros. So now we are able to derive some corollaries: Corollary 2. Let f again be given by a nonzero power series

X n f(X) = anX , n≥0 that converges in Zp. Say α1, . . . , αm are all the zeros of f(X) in Zp. Then there exists a power series g(X) converging in Zp with no zeros in Zp and it holds that: f(X) = (X − α1) ··· (X − αm)g(X). Proof. From the proof of Strassman’s Theorem we know that we can con- struct g1(X) a power series in Zp [[X]] with at most m − 1 zeros and

f(X) = (X − α1)g1(X).

Applying the same procedure to g1 we get g2 with at most m − 2 zeros and

f(X) = (X − α1)(X − α2)g2(X).

Repeating this process we can ﬁnd gm with no zeros and

f(X) = (X − α1) ··· (X − αm)gm(X).

Setting g := gm ﬁnishes the proof.

17 P n Corollary 3. Let f(X) = n≥0 anX be a nonzero power series converging m m in p Zp for some m ∈ Z. Then f(X) has ﬁnitely many zeros in p Zp. The number of zeros is bounded from above by N, where N satiﬁes

mN mn mn mN |p aN |p = max |p an|p and |p an|p < |p aN |p ∀n > N. n≥0

m P mn n m Proof. Deﬁne g(X) := f(p X) = n≥0 anp X . As f converges in p Zp, g converges in Zp. We show that the number of zeros of g in Zp is the same as m m the number of zeros of f in p Zp: If x ∈ Zp is a zero of g, then p x is a zero m m −m of f in p Zp. Conversely if y ∈ p Zp is a zero of f, then p y is a zero of g in Zp. If we apply Strassman to g and the result follows immediately. P n P n Corollary 4. Let f(X) = n≥0 anX and g(X) = n≥0 bnX two p-adic m nonzero power series converging in p Zp. Assume there exist inﬁnitely many m numbers α ∈ p Zp such that f(α) = g(α), then an = bn for all n ∈ N0.

m Proof. Define h(X) = f(X) − g(X) converging in p Zp. If h were nonzero, m it would have a finite number of zeros in p Zp by Corollary 3. But it has m infinitely many zeros in p Zp as all the α’s are zeros of h. So h has to represent the zero power series. Therefore we get an = bn for all n ∈ N0. Remark. Corollaries 2 and 4 have a real respectively complex counterpart.

P n m Corollary 5. Let f(X) = n≥0 anX be nonzero and convergent in p Zp. m If f(x) is periodic, i.e., there exists τ ∈ p Zp such that f(x + τ) = f(x) for m all x ∈ p Zp, then f(X) is constant. Remark. This is not at all what we expect and know in real analysis. For example we know that sine and cosine are periodic power series converging everywhere on R but are not at all constant. The diﬀerence comes from the fact that for power series in R if τ is a period, then nτ for n ∈ Z do not belong to a bounded interval (but in p-adic analysis they do).

Proof. Set h(X) = f(X) − f(0). h has zeros at τn for all n ∈ Z because f is m m m periodic. Since τ ∈ p Zp and p Zp is an ideal, we have that nτ ∈ p Zp. So m h has inﬁnitely many zeros in p Zp. By Corollary 4 this leads us to h(X) = 0 and thus f(X) = f(0) = const.

P n Corollary 6. Let f(X) = n≥0 anX be a nonzero power series converging for all x ∈ Qp. Then f(x) has at most a countable set of zeros. If the set of n→∞ zeros {zn}n≥1 is not ﬁnite, then it forms a sequence with |zn|p −→ ∞.

18 Proof. We know from the second and third talk that if x ∈ Qp then there m exists m ∈ Z such that x ∈ p Zp. So we get that

[ m {x ∈ Qp | f(x) = 0} ⊆ {x ∈ p Zp | f(x) = 0}. m∈Z

m From Corollary 3 we know that in p Zp, f has only finitely many zeros. So the set of all zeros in Qp is a countable union of finite sets, thus countable. Let the set of zeros not be finite and assume that supn≥1 |zn|p ≤ C < ∞. m Then there exists m ∈ Z such that {zn}n≥1 ⊆ p Zp and by Corollary 3 n→∞ {zn}n≥1 is finite, which is the desired contradiction. So |zn|p −→ ∞.