Now I have $V={(a_1,b_1),...,(a_n,b_n)}\subset K^2$, $K$ a field. And I have Lagrange polynomial defined as
$$h(x)=\sum_{i=1}^n b_i \prod_{j\neq i}\frac{x-a_j}{a_i-a_j}$$
and a polynomial f defined as
$$f(x)=\prod_{i=1}^n(x-a_i)$$
Now I need to prove that $I(V)=\big \lbrace \: p\in K[x,y] \mid \forall (x,y)\in V : p(x,y)=0\: \big \rbrace= \langle f(x),y-h(x) \rangle$
I also have a hint that I should divide $g\in I(V)$ by $(f(x),y-h(x))$. But I don't understand it at all, what a result is expected after this divide so that I can prove the statement? I think if I want to prove $(f(x),y-h(x))$ is a Groebner basis, I must divide $(f(x),y-h(x))$ by themselves.
Any hint would be appreciated!
I believe $\langle f(x),y-h(x)\rangle \subseteq I(V)$ is obvious, so let's focus on the other inclusion. Dividing $g\in I(V)$ by $(f(x),y-h(x))$ gives us polynomials $p,q,r\in K[x,y]$ with $g = pf + q(y-h) + r$. Furthermore, for all $(a,b)$,
$$g(a,b) = p(a,b)f(a) + q(a,b)(b-h(a)) + r(a,b) \implies r(a,b) = 0$$
Thus, $r$ possesses at least $n$ different roots, assuming $a_i \neq a_j$ for $i\neq j$, as is usually required when working with interpolation. One can easily prove from here on, that $r=0$ follows. I will leave this part to you, unless you want me to go into further detail. Then $g = pf + q(y-h)$ and the other inclusion is proven.