Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$?
I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery).
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$\begingroup$ Do you want an example over any particular field or over all fields? $\endgroup$– Qiaochu YuanCommented Apr 14, 2010 at 3:07
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$\begingroup$ I was hoping for a list of examples that covered all fields (so it'd be nice if you could add a positive characteristic example). $\endgroup$– jlkCommented Apr 14, 2010 at 3:34
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6 Answers
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If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any purported algebraic relation satisfied by $e^t$.
Edit: Let me fill in a few details. Given a polynomial $P$ in $e^t$ of degree $d$ where each coefficient is a polynomial in $k[t]$ of degree at most $m$, the possible terms that appear in any formal derivative of $P$ lie in a vector space of dimension $(m+1)(d+1)$, so by taking at least $(m+1)(d+1)$ formal derivatives we obtain too many linear relationships between the terms $t^k e^{nt}$. The coefficient of $e^{dt}$ in particular eventually dominates all other coefficients.
answered Apr 14, 2010 at 3:10
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$\begingroup$ I think see your argument. If you write down some more details, then I'd be happy to accept your answer. $\endgroup$– jlkCommented Apr 14, 2010 at 3:25
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7$\begingroup$ jlk: You should accept this answer as it stands. It is not hard to fill in the argument. $\endgroup$ Commented Apr 14, 2010 at 5:05
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$\begingroup$ Only just saw this today. Love it -- so simple and obvious and lovely. $\endgroup$ Commented Sep 2, 2012 at 4:35
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Eisenstein proved (actually, stated) in 1852 that if $f=\sum a_n z^n$ is an algebraic power series with rational coefficients, there exist positive integers $A$ and $B$ such that $A a_n B^n$ are integers for all $n$. In particular, as Eisenstein himself remarks, only finitely many prime numbers appear in the denominators of the coefficients of $f$. For example, $e^z$, $\log(1+z)$, etc., are transcendental.
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$\begingroup$ Can not help saying it is so beautiful $\endgroup$ Commented May 1, 2013 at 8:01
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$\begingroup$ Can you give a reference for that statement? $\endgroup$– user5402Commented Sep 12, 2015 at 16:59
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1$\begingroup$ The original reference is Eisenstein's letter “Über eine allgemeine Eigenschaft der Reihen-Entwicklungen aller algebraischen Functionen”, to be found on p. 765—767 of his Mathematische gesammelte Werke, volume 2 (Chelsea). Other, more modern, references exist. There is for example a proof in Hindry, Silverman's Diophantine Geometry (§E.9). $\endgroup$– ACLCommented Sep 13, 2015 at 19:29
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How about $\sum t^{n!}$? Doesn't a "sea-of-zeroes" argument show it can't be algebraic?
answered Apr 14, 2010 at 3:22
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8$\begingroup$ +1: For the colorful terminology. $\endgroup$ Commented Apr 14, 2010 at 3:24
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15$\begingroup$ Well, that is classically called a lacunary series (from lacus, lake)... I guess classical authors regarded factorial-sized holes as less-than-sea-wide :) $\endgroup$ Commented Apr 14, 2010 at 3:39
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2$\begingroup$ @jlk, more details: let $x$ be the series, let $n$ be a positive integer (the degree of the poynomial hypothetically satisfied by $x$), and consider $1, x, x^2, \dots, x^n$. For $r$ sufficiently large, the coefficient of $t^{r!n}$ will be one in $x^n$ but zero in the smaller powers of $x$. $\endgroup$ Commented Apr 14, 2010 at 4:44
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1$\begingroup$ @Mariano, some seas (e.g., the Sea of Marmara) are smaller than some lakes. $\endgroup$ Commented Apr 14, 2010 at 6:03
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6$\begingroup$ German "See" = English "lake" anyway $\endgroup$ Commented Apr 14, 2010 at 12:57
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Coming back to the lacunary series, I would prefer the series $f(z)=\sum_{k\ge0}z^{d^k}$, where $d>1$ is an integer, because it is the classical example in Mahler's method; this function satisfies the functional equation $f(z^d)=f(z)-z$. I simply copy Ku.Nishioka's argument from her book "Mahler functions and transcendence" (Theorem 1.1.2). Assume that $f(z)$ is algebraic over $\mathbb C(z)$, hence satisfies an \emph{irreducible} equation $f(z)^n+a_{n-1}(z)f(z)^{n-1}+\dots+a_0(z)=0$ where the coefficients $a_j(z)\in\mathbb C(z)$. Substituting $z^d$ for $z$ and using $f(z^d)=f(z)-z$ we obtain $f(z)^n+(-nz+a_{n-1}(z^d))f(z)^{n-1}+\dots=0$. The left-hand sides of both polynomial relations for $f(z)$ must coincide because of the irreducibility. This in particular implies that $a_{n-1}(z)=-nz+a_{n-1}(z^d)$. Letting $a_{n-1}(z)=a(z)/b(z)$ where $a$ and $b$ are two coprime polynomials we see that $a(z)b(z^d)=-nzb(z)b(z^d)+a(z^d)b(z)$. Since $a(z^d)$ and $b(z^d)$ are coprime, $b(z^d)$ must divide $b(z)$. This is possible if only $\deg b(z)=0$, that is, $b(z)=b$ is a nonzero constant. Then $a(z)=-bnz+a(z^d)$ and comparing the degrees of both sides we see that $a(z)$ is constant as well, hence $nz=0$, a contradiction.
answered Apr 16, 2010 at 11:26
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$\begingroup$ IMHO, the factorial series gives a much simpler proof. All you need there is that $nN!$ can be represented as a sum of not more than $n$ factorials in just one way and the next $D$ numbers have no such representation at all if $N>n+D$, say. So the coefficient at the highest power just appears explicitly in the Taylor series of $\sum_{k=0}^n p_kf^k$ $\endgroup$– fedjaCommented Apr 16, 2010 at 14:58
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3$\begingroup$ The factorial series is even simpler if you just refer to the fact that its value at $1/2$ is transcendental due to Liouville's 1844 theorem. Note that Liouville's theorem does not cover $\sum_{k\ge0}z^{d^k}$, so Mahler's example is a bit harder and gives some hints about the powerful method in transcendence number theory. It's a matter of personal taste to choose between "Liouville" and "Mahler". $\endgroup$ Commented Apr 16, 2010 at 23:11
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$\begingroup$ Do you use anywhere in this argument that the base field is C? $\endgroup$ Commented May 9, 2010 at 3:58
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$\begingroup$ If a function $f(z)=\sum_{n=0}^\infty a_nz^n$ with all $a_n\in\mathbb Q$ is algebraic, then it is not hard to show that an algebraic relation can be given over $\mathbb Q(z)$ rather than $\mathbb C$. Therefore the above proof also shows that the formal power series in question is transcendental not only over $\mathbb C(z)$ but over any $k(z)$ where the field $k$ containts $\mathbb Q$. $\endgroup$ Commented May 9, 2010 at 7:57
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- For characteristic 0, http://arxiv.org/abs/1003.2221
- For characteristic p, http://arxiv.org/abs/0810.3709
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$\begingroup$ Ah! I noticed only now that you also want elementary proofs. Well, I don't think the papers cited above use big machinery, but that's subjective of course. $\endgroup$– maksCommented Apr 14, 2010 at 4:34
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over the rationals every power serie with integer coefficents not periodic is tracendent, over Fp a power serie is algebraic iff the secuence of coeficient is p automatic there is a article od jp allouch tracendence of formal series with it information.
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$\begingroup$ For a best example you can see that if $y(t)=\sum a_it^i$ and $f(t,y)=0$ with $f$ a polinomial in two variables you can see the set of $x$ such that $y(x)=1$ is the set of $t$ with $f(t,1)=0$ that is finite, so every y(x) with this set infinite isn´t algebraic particularly $y(t)=e^t$ $\endgroup$– camiloCommented Sep 1, 2012 at 21:37
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3$\begingroup$ Certainly $\frac1{(1-t)^2}=\sum_i(i+1)t^i$ is not transcendental over $\mathbf Q[t]$, is it?? (And you can type dollars to get math; the converse unfortunately doesn't work;-) $\endgroup$ Commented Sep 1, 2012 at 21:42
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1$\begingroup$ this serie is trivially periodic, and dont have integer coeficent . $\endgroup$– camiloCommented Sep 1, 2012 at 21:46
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1$\begingroup$ Your claims are dramatically wrong, or else you are using many conventional words with seriously unconventional meanings. $\endgroup$ Commented Sep 1, 2012 at 23:57
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2$\begingroup$ @camilo: the series of Marc is not periodic and has integer coefficients. How can you claim not integer coefficients?! Are you not understanding the notation? Maybe you think his $i$ stands for the square root of minus 1, and totally misunderstood?? $\endgroup$ Commented Sep 2, 2012 at 4:46