If $V = \emptyset$, what does the ideal $I(V)$ consist of? I'm reading Cox, Little and O'Shea's Ideals, Varieties, and Algorithms.
All my algebra background is one-semester undergraduate abstract algebra.
In the book mentioned above, if $V\subset k^n$ is an affine variety, then the definition of $I(V)$ is $$I(V) = \{f \in k[x_1,..., x_n] : f (a_1,...,a_n) = 0\ for\ all\ (a_1,...,a_n) \in V \},$$ where $k$ is a field.
My question is, if $V$ is empty, what does $I(V)$ consist of ? Is $I(V)=\emptyset$ ?
 A: 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 $f$.
A: Take any polynomial $f \in k[x_1, \dots, x_n]$, and consider $V(f)$. Since the empty set is a subset of any set, we have $\varnothing \subset V(f)$. Then $f \in \mathcal I(V(f)) \subset \mathcal I(\varnothing)$. So, $k[x_1, \dots, x_n] \subset \mathcal I(\varnothing)$ hence $$\mathcal I(\varnothing) = k[x_1, \dots, x_n].$$
