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Let $K$ be the field of fractions of $\mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]]\subset \mathbb{C}[[z, w]].$ Given a formal power series in $t, f\in \mathbb{C}[[t]],$ is there any simple criterion which will conclude that $f(zw)$ does not belong to $K?$ I suspect that $f(zw)\in K$ if and only if $f(t)$ is a rational function, i.e. belongs to $\mathbb{C}(t).$ I am especially interested in proving that $e^{zw}\notin K.$

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For the particular case you're interested in, wouldn't it suffice to let $z = w$ and show that $e^{2z}$ can't be written as a rational function of $z$? –  Michael Lugo Apr 17 '11 at 18:53
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Zamanjan search to prove that $e^{zw} \neq Q(f_1(z),...,f_k(z),g_1(w),...,g_n(w))$ for all rational function $Q$ and all series $f_1(z),...,f_k(z),g_1(w),...,g_n(w)$. –  user12806 Apr 17 '11 at 19:03
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Presumably the vote to close is because the question in the title is easy as Michael showed. (Homework level for a first course in complex variables.) So the title should be changed! –  Gerald Edgar Apr 17 '11 at 19:24
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@Gerald: I've tried to make the title more related to the question –  Fedor Petrov Apr 17 '11 at 19:33
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1+, very interesting question. I've also worked on that to find explicit examples that the functor from complex manifolds to complex locally ringed spaces does not preserve products. –  Martin Brandenburg Apr 17 '11 at 20:10

2 Answers 2

up vote 11 down vote accepted

Theorem 1. Let $f\in\mathbb{C}[[t]]$. Then $f(zw)\in \mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]]$ if and only if $f\in\mathbb{C}[t].$

Proof. The "if" part is obvious. For the "only if" part assume that $f\notin\mathbb{C}[t]$ but $$f(zw) = \sum_{i=1}^n g_i(z)h_i(w),$$ where $g_i\in\mathbb{C}[[z]]$ and $h_i\in\mathbb{C}[[w]]$. We can assume that $n>0$ is minimal. If $g_i(0)=h_i(0)=0$ for all $i$, then $f(0)=0$ and we can divide both sides by $zw$ in the obvious way. Therefore $g_i(0)\neq 0$ or $h_i(0)\neq 0$ for some $i$. Without loss of generality, $g_1(0)=1$, then $$h_1(w)=f(0)-\sum_{i=2}^n g_i(0)h_i(w),$$ so that $$f(zw)=g_1(z)f(0)+\sum_{i=2}^n\bigl(g_i(z)-g_i(0)g_1(z)\bigr)h_i(w).$$ Taking derivative with respect to $w$ and then dividing by $z$ yields $$f'(zw)=\sum_{i=2}^n\frac{g_i(z)-g_i(0)g_1(z)}{z}h_i'(w).$$ Here $f'\notin\mathbb{C}[t]$, but the fractions are elements of $\mathbb{C}[[z]]$, contradicting the minimality of $n$.

Theorem 2. $e^{zw}\notin K$.

Proof. Assume there are $u_i,f_j\in\mathbb{C}[[z]]$ and $v_i,g_j\in\mathbb{C}[[w]]$ such that $$e^{zw}\sum_{i=1}^m u_i(z)v_i(w)=\sum_{j=1}^n f_j(z)g_j(w),$$ where the sums are nonzero. We can assume that $n>0$ is minimal, and also that $m>0$ is minimal for this $n$. In particular, the $u_i$'s are independent over $\mathbb{C}$. We can assume that some $v_i(0)$ or $g_j(0)$ is nonzero, for otherwise we can divide both sides by $w$ in the obvious way. Then $$ \sum_{i=1}^m u_i(z)v_i(0)=\sum_{j=1}^n f_j(z)g_j(0) $$ is a nonzero element of $\mathbb{C}[[z]]$, call it $h(z)$. In particular, $g_j(0)\neq 0$ for some $j$. Without loss of generality $g_1(0)=1$, then with the notation $$\tilde u_i(z):=u_i(z)/h(z)\in\mathbb{C}((z))$$ $$\tilde f_j(z):=f_j(z)/h(z)\in\mathbb{C}((z))$$ $$e^{zw}\sum_{i=1}^m \tilde u_i(z)v_i(w)=g_1(w)+\sum_{j=2}^n \tilde f_j(z)\bigl(g_j(w)-g_j(0)g_1(w)\bigr).$$ The sum on the right hand side is independent of $z$ for otherwise we could contradict the minimality of $n$ by taking the $z$-derivative on both sides and then multiplying by a large power of $z$ to turn elements of $\mathbb{C}((z))$ into elements of $\mathbb{C}[[z]]$. At any rate, the right hand side is a nonzero element $k\in\mathbb{C}[[w]]$, because $k(0)=g_1(0)=1$. With the notation $$ \tilde v_i(w):=v_i(w)/k(w)\in\mathbb{C}((w))$$ we conclude $$e^{-zw}=\sum_{i=1}^m \tilde u_i(z)\tilde v_i(w).$$ In particular, there is some integer $r>0$ such that $$(zw)^r e^{-zw}\in \mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]],$$ which contradicts Theorem 1. The proof of Theorem 2 is complete.

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Very nice! 1+ –  Martin Brandenburg Apr 18 '11 at 8:13
    
Very nice indeed! Thanks! –  zamanjan Apr 18 '11 at 18:46

If $e^{zw}=\frac{\sum_{i=1}^{a} f_i(z)g_i(w)}{\sum_{j=1}^b u_j(z)v_j(w)}$,

$e^{zw}(\sum u_j(z)v_j(w))=\sum f_i(z)g_i(w)$

$\frac{\partial^{n}}{\partial w^n}e^{zw}(\sum u_j(z)v_j(w))=\frac{\partial^{n}}{\partial w^n}\sum f_i(z)g_i(w)$

$\sum_{k=0}^n C^k_n z^k e^{zw}(\sum u_j(z)v^{(n-k)}_j(w))=\sum f_i(z)g_i^{(n)}(w)$

If $w=0$, $\sum_{k=0}^n C^k_n z^k(\sum u_j(z)v^{(n-k)}_j(0))=\sum f_i(z)g_i^{(n)}(0)$

For $(f_i(z))$ a finite family. $N \in \mathbb{N}$ exists such that, for all $(\lambda_i)$,if $\sum \lambda_i f_i(z) \neq 0$, $valuation( \sum \lambda_i f_i(z)) \leq N$.

Let $h_{k}(z)=\sum_{j=1}^b u_j(z)v_j^{(k)}(0)$. The infinite family $(h_0,...,h_k,...)$ has rank less or equal than $b$.

We have:

$$h_0 \in Vect(f_i)$$ $$zh_0+h_1 \in Vect(f_i)$$ $$z^2h_0+2zh_1 +h_2\in Vect(f_i)$$ $$z^3h_0+3z^2h_1+3zh_2+h_3 \in Vect(f_i)$$

If $h_0=0$, a diagonal is empty.

If $h_1= \lambda h_0$, we can substract $\lambda \times$ the first line to the second line. We can substract $2\lambda \times$ the second line to the third. And substract $k\lambda \times $ the line $k$ to the line $k+1$, etc... A diagonal vanishes.

We can suppose that the family $(h_1,...,h_c)$ generates the infinite family $(h_0,...,h_k,...)$

If $h_d=\lambda_0 h_0 + ... \lambda_c h_c$, we substract to the $(d+i+1)$-th line, $C^i_{d+i}/C^i_i \times \lambda_0 \times$ line $1$-th, ...,$C^i_{d+i}/C^i_{j+i} \times \lambda_j \times$ line $(j+1)$-th...,$C^i_{d+i}/C^i_{c+i}\times \lambda_c \times$ line $(c+1)$-th.

The diagonal $d+1$ vanishes.

So, if $c \neq 0$, the valuation is not bounded in $Vect(f_i)$. Contradiction.

My answer is not complete.

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