I have been reading Fulton's Algebraic curves, chapter 1.3. The chapter starts by proving some properties about the ideal of a set of points for which the polynomials in that ideal vanish. This is denoted $I(X) \subseteq K[x_1, \dots , x_n]$ where $X \subseteq \mathbb A^n(K)$, $K$ a field.
Then I started working on the following question ($1.17$) and I feel a bit uncertain:
(a) Let $V$ be an algebraic set in $\mathbb A ^n (K)$. Let $P \in \mathbb A^n(K) \setminus V$ be a point not in $V$. Show that there is a polynomial $F \in K[X_1, \dots, X_n]$ such that $F(Q)=0$ for all $Q\in V$, but $F(P)=1$.
(Hint: $I(V) \neq I(V \cup \{P\})$
The hint seems to suggest that I take some polynomial that vanishes on $V$, but not on $P$. Which is indeed what I want. So we should take some $G\in I(V)$ such that $G \not \in I(V \cup \{P\}) $. Since $G$ does not vanish at $P$, we know that $G(P) \neq 0$. Is it really as simple as now taking $$ F(X_1, \dots, X_n):=\frac{1}{G(P)}G(X_1, \dots, X_n) \quad ?$$
(b) Let $P_1, \dots ,P_r$ be distinct points in $\mathbb A^n(K)$, not in an algebraic set $V$. Show that there are polynomials $F_1, \dots, F_r \in I (V)$ such that $F_i (P_j )= 0$ if $i \neq j$ , and $F_i (P_i ) = 1$.
(Hint: Apply (a) to the union of V and all but one point.)
Here we generalise the result from question (a) to form an orthonormal basis of polynomials. My gut feeling is screaming " use induction on the amount of points $r$", but this is not what the hint is saying, I think.
Base case: $r=1$ this follows from (a) since indeed $F(P)=1$, but there are no other points to check so the other condition holds trivially.
Induction Hypothesis Suppose now that there are distinct points $P_1, \dots, P_m$ for some $m\in \mathbb N$, such that for $1\leq i,j \leq m$ there are $F_1, \dots, F_m \in I (V)$ with $F_i (P_j )= 0$ if $i \neq j$ , while $F_i (P_i ) = 1$.
We will now make the inductive step Now suppose that there are distinct points $P_1, \dots, P_{m+1}$. We now need to construct a final polynomial that evaluates to zero for all the prior points and to $1$ for $P_{m+1}$. I am stumped. Maybe I should instead use the hint, but I do not know how.
(c) With $P_1, \dots ,P_r$ and $V$ as in (b), and $a_{ij}\in K$ for $1 \leq i , j \leq r$ , show that there are $G_i \in I (V)$ with $G_i (P_j )=a_{ij}$ for all $i$ and $j$ . (Hint: Consider $\sum_{j}a_{ij}F_j$ .)
For this question we consider, as the hint suggests: $$ G_i= \sum_{k=1}^ra_{ik} F_{k}$$ Where question (b) gives us that $F_1, \dots, F_r \in I (V)$ such that $F_k (P_j )= 0$ if $k \neq j$ , and $F_k (P_k ) = 1$.
Observe then that $$ G_i(P_j):=\sum_{k=1}^r a_{ik}F_k(P_j)= a_{ij} $$ Since the only nonzero term in the sum is where $k=j$.
EDIT. as pointed out in the comments, I tricked myself with the summation variable. I changed the indices.
My questions have been answered in the comments:
(a) is fine
(b) the finite union of algebraic sets is an again an algebraic set. Since points are algebraic sets we can take the union over all but one point $P_i$ with $V$ and apply (a), this gives the desired polynomial $F_i$ for each point $P_i$.
(c) Is fine as it is now, but I had made a summation error by plugging in $P_j$ while $j$ was also the summation variable.