In page 16 of these notes on $p$-adic $L$-functions, it makes the following claim:

Let $\alpha$ be a $p$-adic measure on $\mathbf{Z}_p$ which corresponds to a power series $F_{\alpha}(T) \in \mathbf{Z}_p[[T]]$ under the Iwasawa isomorphism. Fix any $a \in \mathbf{Z}_p^{\times}$. Then the measure $\alpha \circ (a)$ corresponds to the power series $F_{\alpha}((1+T)^{a}-1) \in \mathbf{Z}_p[[T]]$ under the Iwasawa isomorphism.

My question is: what exactly does $F_{\alpha}((1+T)^{a}-1)$ mean? That is, for a general $a \in \mathbf{Z}_p^{\times}$, the expression $(1+T)^{a}-1$ is a power series (not necessarily a polynomial). So when we write $F_{\alpha}((1+T)^{a}-1)$, we are plugging in a power series into another power series.

Does it make sense to do this? Even once we expand out all the brackets and group all like terms, wouldn't we still have to add infinitely many coefficients for each $T^n$ term? So we'd have to worry about convergence issues for the coefficients. Is it known that the coefficients for each sum would always converge?

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    $\begingroup$ Are you sure it isn't $(1+T)^a-1$? $\endgroup$
    – efs
    May 12 at 4:56
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    $\begingroup$ Anyway, you can always formally take the composition $F(G(T))$ of two power series $F(T)$ and $G(T)=T+\cdots$. $\endgroup$
    – efs
    May 12 at 5:01
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    $\begingroup$ The text you quote does not appear on p. 16, nor on the neighboring pages. $\endgroup$
    – abx
    May 12 at 5:06
  • $\begingroup$ You don't say in your question what $\alpha \circ (a)$ means. Perhaps the result you ask about is, up to some normalization, essentially Theorem 1.2 in Chapter 4 of Lang's Cyclotomic Fields I and II. $\endgroup$
    – KConrad
    May 12 at 6:07
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    $\begingroup$ You still have a typo, writing $(1 + T^a) - 1$, which is just $T^a$, instead of $(1+T)^a-1$. The expression $T^a$ for general $a \in \mathbf Z_p$ is meaningless, but $(1+T)^a - 1 = \sum_{n \geq 1} \binom{a}{n}T^n$ is a legitimate power series in $\mathbf Z_p[[T]]$ for $a \in \mathbf Z_p$. $\endgroup$
    – KConrad
    May 12 at 14:38

1 Answer 1


You are asking about substitution of one power series into another. This can be interpreted in two ways, which ultimately amount to the same thing (like two different ways of thinking about anything) but one method is elementary and rather clunky while the other takes longer to set up but ultimately is more slick and better for actually proving properties in a non-tedious way.

Method 1: the combinatorial method (using finitely many terms at a time). If $f(x) = a_0 + a_1x + a_2x^2 + \cdots$ is an element of $A[[x]]$, where $A$ is a commutative ring, and $g(x) = b_1x + b_2x^2 + \cdots$ is another element of $A[[x]]$ with constant term $0$, then $f(g(x))$ makes sense as $$ a_0 + a_1g(x) + a_2g(x)^2 + \cdots $$ because $g(x)^n = b_1^nx^n + \cdots$ has no terms in degree below $n$, so to find the coefficient of $x^d$ in $f(g(x))$ you only need to work with the finite sum $a_0 + a_1g(x) + \cdots + a_{d}g(x)^{d}$ and extract the coefficient of $x^d$ in that sum. It "doesn't make sense" to compose $f(g(x))$ if $g(x) = b_0 + b_1x + b_2x^2 + \cdots$ has a nonzero constant term because the constant term of $f(g(x))$ would then be $f(g(0)) = f(b_0) = a_0 + a_1b_0 + a_2b_0^2 + \cdots$ and this is an infinite series that doesn't make algebraic sense unless $b_0$ is nilpotent or some other reason causes all but finitely many terms to be $0$.

Example: if $f(x) \in \mathbf Z_p[[x]]$ then $f((1+x)^a - 1)$ makes sense since $(1+x)^a - 1 = \sum_{n \geq 1} \binom{a}{n}x^n$ is a power series with constant term $0$.

I call this the combinatorial method because I have seen combinatorialists describe computations with formal power series in this way, which is purely algebraic and emphasizes the finitely many calculations needed to work out any particular coefficient. It seems like a clunky way of setting up formal power series if you want to prove properties of them beyond the simplest ones, but it has the advantage of being intuitive and getting to the point of things in a direct way. To prove things from this point of view you typically need to work them out degree-by-degree with induction. See, for example, how proofs look in the Amer. Math. Monthly article on formal power series by Ivan Niven here.

Now let's try something "completely different" that is more in the $p$-adic spirit.

Method 2: the $x$-adic topology. For a commutative ring $A$, define the $x$-adic absolute value on $A[[x]]$ by $|f(x)|_x = (1/2)^n$ when $f(x) = c_nx^n + \cdots$ is nonzero with its first nonzero term occurring in degree $n$, and $|0|_x = 0$. For example, all nonzero $a \in A$ have $|a|_x = 1$ when we view $a$ in $A[[x]]$ as a constant power series.

For all $f$ and $g$ in $A[[x]]$, $|f+g|_x \leq \max(|f|_x,|g|_x)$ and $|fg|_x \leq |f|_x|g|_x$. If $A$ is an integral domain, such as $A = \mathbf Z_p$ or $\mathbf Q_p$, then $|fg|_x = |f|_x|g|_x$. From $|\cdot|_x$ we get a topology on $A[[x]]$ in which $x^n \to 0$ as $n \to \infty$, and $|f|_x < 1$ if and only if $f$ has constant term $0$. The $x$-adic distance between $f$ and $g$ is defined to be $|f-g|_x$, which is a non-Archimedean metric on $A[[x]]$ that makes it complete. Just as in the $p$-adic numbers, an sequence $\{f_n\}$ in $A[[x]]$ is $x$-adically convergent if and and only if $|f_{n+1} - f_n|_x \to 0$ as $n \to \infty$, so an infinite series of elements in $A[[x]]$ is $x$-adically convergent if and only if its general term tends to $0$.

Now let's talk about the composition $f(g(x))$ for $f(x) \in A[[x]]$ and $g(x) \in A[[x]]$. What does it mean and when does it make sense? To take advantage of $x$-adic convergence, let's view $f(x)$ as a power series in a different indeterminate, say $f(y) \in R[[y]]$ where $R = A[[x]]$ is a non-Archimedean complete ring and $f(y)$ happens to have all "constant" coefficients (lying in $A$, not just in $R$). For example, if $f(x) = 1 + x + x^2 + x^3 + \cdots$ then we view it as $1 + y + y^2 + y^3 + \cdots$ in $R[[y]]$ whose coefficients are all $1 \in A$. Asking about the meaning of $f(g(x))$ is asking about where a power series in $R[[y]]$ converges on $R$. Just like a $p$-adic power series in $\mathbf Z_p[[y]]$ has a $p$-adic disc of convergence, a power series $f(y)$ in $R[[y]]$ has a disc of convergence in $R$: if $f(y) = \sum_{n \geq 0} c_ny^n$, then $f(r)$ converges if and only if $|c_nr^n|_x \to 0$ as $n \to \infty$. Since $|c_nr^n|_x \leq |c_n|_x|r|_x^n \leq |r|_x^n$, $f(r)$ converges if $|r|_x < 1$. Since $r^n - s^n$ is divisible by $r-s$ for all $n \geq 0$, $|f(r) - f(s)|_x \leq |r-s|_x$ when $f(r)$ and $f(s)$ converge, so the power series $f(y)$ is $x$-adically uniformly continuous on its disc of convergence.

Writing $r \in R = A[[x]]$ as $g(x)$, saying $|r|_x < 1$ means $g(x)$ has constant term $0$. This shows $f(g(x))$ converges $x$-adically in $A[[x]]$ when $g(x)$ has constant term $0$. It's left to you to check, using the $x$-adic uniform continuity estimate $|f(g(x)) - f(h(x))|_x \leq |g(x) - h(x)|_x$ when $g(x)$ and $h(x)$ have constant term $0$, that the calculation of the coefficients of $f(g(x))$ as an $x$-adic limit (finding each of its coefficients as an element of $A[[x]]$) can be done by the combinatorial method ("finitely many terms at a time") because $g(x) \equiv g_d(x) \bmod x^{d+1}$ where $g_d(x)$ is the truncation of $g(x)$ to the sum of its terms up through degree $d$.

Example: if $f(y) \in \mathbf Z_p[[y]] \subset (\mathbf Z_p[[x]])[[y]]$ then $f((1+x)^a - 1)$ makes sense in $\mathbf Z_p[[x]]$ since $|(1+x)^a - 1|_x < 1$, and that's because $(1+x)^a - 1 = \sum_{n \geq 1} \binom{a}{n}x^n$ has constant term $0$.

If $A$ is an integral domain (like $\mathbf Z_p$ or $\mathbf Q_p$) and $f(y) = \sum_{n \geq 0} c_ny^n$ is a genuine power series in $A[[y]]$ -- it has infinitely many nonzero coefficients in $A$ -- then $|c_nr^n|_x = |c_n|_x|r|_x^n = |r|_x^n$ infinitely often (whenever $c_n \not= 0$), and this tends to $0$ only if $|r|_x < 1$, so $f(r)$ converges if and only if $|r|_x < 1$. That means $f(g(x))$ converges $x$-adically in $A[[x]]$ if and only if $g(x) \in A[[x]]$ has constant term $0$.

Remark 1. Armed with the $x$-adic topology, we gain access to Hensel's lemma and topological arguments that can replace tedious degree-by-degree arguments in the combinatorial method. (It is not lost on me that the proof of Hensel's lemma itself is based on a degree-by-degree argument, so my point of comparison between Methods 1 and 2 is that Method 1 uses degree-by-degree arguments very often, while Method 2 can often avoid them with simple continuity arguments.) For example, the formal derivative on $A[[x]]$ is $x$-adically uniformly continuous (since $|f'|_x \leq (1/2)|f|_x$), so to prove that the usual rules of formal derivatives are valid on $A[[x]]$, by its continuity and the denseness of $A[[x]]$ we are reduced to verifying the rules on polynomials, where the rules are "already known". There is no need to prove the identities on actual power series, but you may need to be more careful in the case of proving the chain rule to reduce to the polynomial case.

Remark 2. Above I am not saying $f(g(x))$ can never make sense in $A[[x]]$ if $f(x)$ is a genuine power series (not a polynomial) and $g(x)$ has a nonzero constant term, but only that it does not make $x$-adic sense (if $A$ is an integral domain). If $A = \mathbf Z_p$, so $A$ itself has a nontrivial notion of convergence, then you may be able to evaluate $f(g(x))$ when $g(x)$ has a nonzero constant term $b_0$ such that $f(b_0)$ converges in $A$. For example, if $f(y) = 1 + y + y^2 + y^3 + \cdots$ has all coefficients equal to $1$ then $f(p+x)$ makes sense in $\mathbf Z_p[[x]]$ but not with the $x$-adic topology since $|p+x|_x = 1$. You'd have to use something like the $(p,x)$-adic topology instead, where elements of $\mathbf Z_p[[x]]$ are small not just if they start off with a large power of $x$, but if they are in a high power of the ideal $(p,x)$, meaning below some high power of $x$ the coefficients are very divisible by $p$. Since $(x) \subset (p,x)$ we have $(x)^n \subset (p,x)^n$, so $x$-adically small power series in $\mathbf Z_p[[x]]$ are also $(p,x)$-adically small, but not conversely (e.g., $p^n+x \to x$ in the $(p,x)$-adic topology but not in the $x$-adic topology).

  • $\begingroup$ thanks a ton! this second way of thinking about power series composition is very illuminating. I appreciate the help :) $\endgroup$ May 12 at 19:31
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    $\begingroup$ It's a good exercise to go through that Niven paper on formal power series and prove the results using Hensel's lemma and the $x$-adic topology in place of degree-by-degree arguments, in his $\mathbf C[[x]]$ (which he writes as $P$) or in $A[[x]]$ when possible (sometimes $A$ has to be a field of characteristic $0$). For example, Theorem 1 in $A[[x]]$ is saying $A[[x]]^\times$ is the series with constant term in $A^\times$ (use Hensel's lemma) and Theorem 3 is saying each $\alpha \in 1 + xA[[x]]$ has a unique $n$th root in $1+xA[[x]]$ if $n \in A^\times$ (Hensel's lemma on $y^n - \alpha$). $\endgroup$
    – KConrad
    May 12 at 20:13
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    $\begingroup$ My comments about degree-by-degree arguments now reminds me of Lubin's description of the development of Lubin-Tate theory on the MO page mathoverflow.net/questions/220796/motivating-lubin-tate-theory, where he writes "I used extremely tiresome degree-by-degree methods based on the techniques of Lazard" until he made a discovery that led Tate to the important Lemma 1 of their 1965 Annals paper, which codifies many degree-by-degree arguments into a single setting so you don't have to go through that actual process directly all the time. $\endgroup$
    – KConrad
    May 12 at 20:21

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