If $V = \emptyset$, what does the ideal $I(V)$ consist of?

Solution 1:

No, to the contrary. The ideal $I(V)$ is the set of all polynomials $f$ such that for each element $(a_1, \dots, a_n) \in V$, you have $f(a_1, \dots, a_n) = 0$. But all the polynomials satisfy this condition, because there are not elements $(a_1, \dots, a_n)$ to check! So $I(V) = k[x_1, \dots, x_n]$.

This is called a vacuous truth. A statement of the form "For all elements $x$ of the empty set, condition $P(x)$ is true" are always true statements, because there are no elements $x$ to check.

Maybe it will help to rephrase a bit. Suppose $n=1$ for simplicity. You're asking "What are the polynomials $f \in k[x]$ such that the set of roots of $f$ contains $V$?" But if $V = \varnothing$ then all the polynomials work, because whatever $f$ is, then $\varnothing$ will be included in the set of roots of $f$.