K.N wrote: Do there exists many infinitely points like $(x_1,y_1),(x_2,y_2),...$ such that for any sequences like {$b_1,b_2,...$} of real numbers there exists a polynomial $P(x,y)\in R[x,y]$ such that we have for all $i$ : $P(x_{i},y_{i})=b_{i}$ No : We need $(x_i,y_i)\ne(x_j, y_j)$ (else choosing $b_i=b_j$ leads to impossibility). Then, choosing $b_1=1$ and $b_i=0$ $\forall i>1$ leads to impossibility

pco wrote: K.N wrote: Do there exists many infinitely points like $(x_1,y_1),(x_2,y_2),...$ such that for any sequences like {$b_1,b_2,...$} of real numbers there exists a polynomial $P(x,y)\in R[x,y]$ such that we have for all $i$ : $P(x_{i},y_{i})=b_{i}$ No : We need $(x_i,y_i)\ne(x_j, y_j)$ (else choosing $b_i=b_j$ leads to impossibility). Then, choosing $b_1=1$ and $b_i=0$ $\forall i>1$ leads to impossibility What's impossible?

bgn wrote: What's impossible? Because this would mean infinitely many zeroes for a nonzero polynomial

pco wrote: choosing $b_1=1$ and $b_i=0$ $\forall i>1$ leads to impossibility Actually, the points: $(x_{1},y_{1})=(1,0)$ and $(x_{i},x_{i}) \forall i>1$ and $P(x,y)=x-y$ work just fine with your sequence.

pco wrote: bgn wrote: What's impossible? Because this would mean infinitely many zeroes for a nonzero polynomial A two variable polynomial with infinitely many roots is not necessarily zero, like $P(x,y)=x-y$

You are quite right ! Sorry for my error. And thanks.

Suppose for contradiction that these points exist. Take a sequence with $b_i = 1$ and $b_j = 0, \forall j \neq i$, we obtain a polynomial $P_i$ such that $P_i(x_i, y_i) = 1$ and $P_i(x_j, y_j) = 0, \forall j \neq i$. Consider $P_1, P_2$, and let $Q \in \mathbb{R}[x,y]$ be their greatest common divisor. By Bezout's theorem, $\frac{P_1}{Q}$ and $\frac{P_2}{Q}$ vanish simultaneously at finitely many points. But $P_1$ and $P_2$ vanish at $(x_i, y_i)$ for any $i > 2$. Thus there exists a positive integer $m > 2$ such that $Q(x_i, y_i) = 0$ whenever $i \geq m$. Next let $R$ be the greatest common divisor of $Q$ and $P_m$. The same argument yields a positive integer $n > m$ such that $R(x_i, y_i) = 0$ whenever $i \geq n$. Continuing this process, we obtain a sequence of polynomials $P_1, Q, R, \dots$ that are distinct, and every polynomial in the sequence divides its previous term. That is just impossible!

pco wrote: You are quite right ! Sorry for my error. And thanks. My pleasure.

any solution?

already above.

Take a converging sequence $b_i$ that converges to $b$ after $i\ge k$ and let $b_1\neq b_2\neq ....\neq b_k$ So u find finite many pairs $(x_1,y_1),(x_2,y_2),....(x_k,y_k)$ then after $b_k$ all values of the seq. are $b$. Then for $b_{(i\ge k)}$ if u find same pairs $(x_k,y_k)$ then done. If not u then u find distinct pairs for which $P(x_i,y_i)=b$ so that means the $P(x,y)$ remains constant for $x\to \infty$ and $y\to \infty$ or one of them $\to \infty$.So its a contradiction!

angiland wrote: Consider $P_1, P_2$, and let $Q \in \mathbb{R}[x,y]$ be their greatest common divisor. By Bezout's theorem I can see how $P_1$ and $P_2$ can have a GCD, even though $R[X,Y]$ is not a PID, but how can Bezout's theorem work in this case?

K.N wrote: Do there exists many infinitely points like $(x_1,y_1),(x_2,y_2),...$ such that for any sequences like {$b_1,b_2,...$} of real numbers there exists a polynomial $P(x,y)\in R[x,y]$ such that we have for all $i$ : $P(x_{i},y_{i})=b_{i}$ No, such points do not exist. The idea is simple. Fixing some polynomial, it cannot increase too fast. So, if we choose $b_1,b_2,\dots$ to increase, say exponentially, there is no chance a polynomial to interpolate these values. The details follow. For the sake of contradiction, suppose such sequence $(x_i,y_i)_{i\in\mathbb{N}}$ does exist. If it's bounded, it's enough to choose an unbounded sequence $(b_i)_{i\in\mathbb{N}}$. Since any polynomial $P(x,y)$ is bounded when $(x,y)$ runs over a bounded region, it brings us to a contradiction. Suppose now $(x_i,y_i)_{i\in\mathbb{N}}$ is unbounded sequence. Consider $b_i := e^{1+\max(|x_i|,|y_i|)},i=1,2,\dots$. Assume, there exists a polynomial $P(x,y)$ of degree $n$ with $P(x_i,y_i)=b_i$. Since $|x_i^k y_i^{\ell}|\leq (\max(|x_i|,|y_i|))^{k+\ell}\leq (1+\max(|x_i|,|y_i|))^n$, for any $n\geq k+\ell$, we get: $$b_i=|P(x_i,y_i)|\leq c(1+\max(|x_i|,|y_i|))^n$$where $c>0$ is the sum of absolute values of the coefficients of $P$. Hence: $$e^{z_i}\leq cz_i, i=1,2,\dots$$where $z_i=1+\max(|x_i|,|y_i|)$ and $c$ is a constant (not depending on $i$). But it's a contradiction, since $z_i$ may take values as large as we want.