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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Sign up to join this communityLet $k$ 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).
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.
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.
How about $\sum t^{n!}$? Doesn't a "sea-of-zeroes" argument show it can't be algebraic?
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.
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.