Why $I(\mathbb{A}^n)=(0)$? Let $\mathbb{A}^n$ denote the set of n-tuples of elements from field $k$ and $I(X)$ the ideal of polynomials in $k[x_1,...,x_n]$ that vanish every point in $X$. The note I’m reading, in showing that $\mathbb{A}^n$ is affine variety, says $I(\mathbb{A}^n)=(0)$.
Why is that? I know it’s probably extremely trivial but I have very little background so please bear with me... Suppose that $k$ is $\mathbb{Z}_p$ then something like $x^p-x$ also belongs to $I(\mathbb{Z}_p)$ so I guess k must be infinite too (I don’t see this mentioned in the note)? Or am I confusing something?
 A: Let me just write up a proof without going through a contradiction (it's almost literally the same proof but it doesn't use contradiction so is somewhat cleaner and clearer)
It proceeds by induction, like dcolazin's proof . The proof for $n=1$ is the same. 
Now let $p\in I(\mathbb{A}^{n+1})$, write $p(x_1,...,x_n, y) = \displaystyle\sum_{k=0}^da_k(x_1,...,x_n)y^k$. 
Fix $(x_1,...,x_n)\in \mathbb{A}^n$. 
Then for all $y$, $p(x_1,...,x_n, y)=0$, therefore $\displaystyle\sum_{k=0}^da_k(x_1,...,x_n)X^k$ is the zero polynomial : $a_k(x_1,...,x_n) = 0$ for all $k$. 
But this is for any $(x_1,...,x_n)$, hence for all $k$, $a_k\in I(\mathbb{A}^n)$, hence for all $k, a_k=0$ by induction. Therefore $p=0$.
A: If $k$ is finite, you can build polynomials what vanish in all $\mathbb{A}^n$ thanks to $p(x) = \Pi_{z\in k}(x-z)$.
If $k$ is infinite, then $I(\mathbb{A}^1) = 0$ because polynomials in one variable only have finite solutions. 
Then by induction: suppose $I(\mathbb{A}^n) = 0$. Consider $p\in I(\mathbb{A}^{n+1})$, then $p(x_1,\ldots,x_n,y) = \sum_{i=0}^d a_d(x)y^d$, where $a_d \in k[x]$ and $x = (x_1,\ldots,x_n)$. 
If $d=0$, then $p\in I(\mathbb{A}^{n})$ and you can apply the inductive hypothesis.
Suppose, by contradiction, that $\exists i \in \{0,\ldots,d\}$ such that $a_i(x) \neq 0$. Then by inductive hypothesis $\exists \tilde{x} \in \mathbb{A^n}$ such that $a_i(\tilde{x}) \neq 0$. Then $p(\tilde{x},y) \in I(\mathbb{A^1})$ and $p(\tilde{x},y) \neq 0$, which is absurd.
