TITLE: Formal power series is Taylor expansion of rational function iff Hankel determinants vanish?
QUESTION [9 upvotes]: Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M\gg0$?

REPLY [2 votes]: This is Lemma III, attributed to Kronecker, on page 5 of Salem, Algebraic Numbers and Fourier Analysis. The book has been posted to the web by Plouffe, at http://plouffe.fr/simon/math/Algebraic%20Numbers%20and%20Fourier%20Analysis.pdf

REPLY [2 votes]: I am not going to add anything new, just what is my take on it. The proof consists of four really simple steps. 
Step I. Determinant is zero iff there exist numbers $b_0,\dots,b_N$, not all of them zero, such that 
$$\sum_{i=0}^N a_{M+i+j}b_i=0$$
for all $0\le j\le N$. 
(Proof: Basically, the definition of determinant.)
Step II. Determinant is zero iff there exists a nonzero polynomial
$B(T)=b_0T^N+\dots+b_N$ such that in the series
$$B(T)u(T)=\sum_{n=0}^\infty c_n T^n$$
we have
$$c_{M+N}=c_{M+N+1}=\dots=c_{M+2N}=0.$$
(Proof: Multiplication of series.)
Step III. The series $u(T)$ is rational iff there exists $N>0$ and a nonzero polynomial $B(T)=b_0T^N+\dots+b_N$ such that in the series
$$B(T)u(T)=\sum_{n=0}^\infty c_n T^n$$
we have
$$c_{M+N}=c_{M+N+1}=\dots=c_{M+2N}=0$$
for $M\gg 0$. (Proof: just the definitions!)
Step IV. The series $u(T)$ is rational iff there exists $N>0$ such that the determinant is zero for $M\gg 0$.
Now, I cheated in the step IV: even if the determinant is zero,  the polynomial $B(T)$ may actually depend on $M$, which is wrong! Actually, this is the hard part. The standard way to fix this is to chose the smallest possible $N$, as it is done in Koblitz.