Equivalence of Classical Nullstellensatz to "Affine schemes have points"

Let's show the equivalence between two versions of nullstellensatz:

1. In terms of ideals, the theorem from the book says that $V(I) neq emptyset$ iff $I neq (1)$, where $V(I) = { c in mathbb C^n : forall f in I, f(c) = 0 }$
2. If $mathfrak m$ is a maximal ideal of $C[X_1, dots X_n]$, then there exists $(c_1, c_2, dots c_n) in mathbb C^n$ such that $mathfrak m = (X_1 - c_1, dots X_n - c_n)$. Also, every ideal of the form $(X_1 - c_1, dots, X_n - c_n)$ is maximal.

Which I will write as:

• 2.a If $mathfrak m$ is a maximal ideal of $C[X_1, dots X_n]$, then there exist $c_1, c_2, dots c_n$ such that $mathfrak m = (X_1 - c_1, dots X_n - c_n)$
• 2.b Every ideal of $C[X_1, dots, X_n]$ of the form $(X_1 - c_1, dots, X_n - c_n)$ is maximal.

We can prove (2.b) by induction on the number of variables. When $n = 1$, we have $mathbb C[X]$ where all ideals of $mathbb C[X]$ are of the form $(X - c)$ since it's a principal ideal domain. Also, the quotient ring $mathbb C[X]/(X - c) simeq C$, since we will be left with polynomials of degree $0$ on taking reminders with a degree $1$ polynomial, $(X - c)$. That is, we will be left with $mathbb C$. By induction on $n$, when $n = k + 1$, write the ring $mathbb C[X, X_k, X_k+1]$ as $(mathbb C[X, dots, X_k])/[X_k+1]$. Given some ideal of the form $(X_1 - c_1, dots X_{k+1} - c_{k+1})$, perform the quotienting as:

begin{align*} &(mathbb C[X, dots, X_k, X_{k+1}]/(X_1 - c_1, dots X_k - c_k, X_{k+1} - c_{k+1}) \ &=(mathbb C[X, dots, X_k]/(X_1 - c_1, dots X_k - c_k))[X_{k+1}]/(X_{k+1} - c_{k+1}) quad text{(factor in terms of $X_{k+1}$)}\ &= mathbb C[X_{k+1}]/(X_{k+1} - c_{k+1}) quad text{(Induction hypothesis)} \ &= mathbb C quad text{(Similar to $n = 1$)} end{align*}

So the interest implications are between (1) and (2.a)

(1) implies (2.a):

We know that $V(I) neq emptyset$ iff $I neq (1)$. We wish to show that if $mathfrak m$ is a maximal ideal of $C[X_1, dots, X_n]$, then we have a point $(c_1, c_2, dots, c_n) in mathbb C^n$ such that $mathfrak m = (X_1 - c_1, dots, X_n - c_n)$. The proof proceeds in two stages:

1. Since $mathfrak m neq 1$, we have that $V(mathfrak m) neq emptyset$.
2. We must have that $V(mathfrak m) = { cstar }$ for some $cstar in mathbb C^n$. For contradiction, assume not.
3. Then consider the ideal $J = (X_1 - cstar_1, dots, X_n - cstar_n)$.
4. $V(J) = { cstar_n }$. Clearly, $cstar in V(J)$ since the polynomials $X_i - cstar_i$ vanish at $cstar$. To show that this is the only point: if we have that for some point $d in mathbb C^n; d in V(J)$, then we have that $d_i - cstar_i = 0$. This implies $d = cstar$.
5. Since $V(J) subsetneq V(mathfrak m)$, we have that $mathfrak m subsetneq J$ which contradicts the maximality of $mathfrak m$
6. Hence, we have that every maximal ideal $mathfrak m$ can be written as $mathfrak m = (X_1 - c_1, dots, X_n - c_n)$ for some

(2.a) implies (1)

We know that $mathfrak m$ is a maximal ideal of $C[X_1, dots, X_n]$, then we have a point $(c_1, c_2, dots, c_n) in mathbb C^n$ such that $mathfrak m = (X_1 - c_1, dots, X_n - c_n)$. We wish to show that $V(I) neq emptyset$ iff $I neq (1)$.

Forward: $V(I) neq emptyset implies I neq (1)$:

we have an ideal $I$ such that $V(I) neq emptyset$. This means that we have a point at which all polynomials in $I$ evaluate to $0$. But $1$ never evaluates to $0$. Hence $1 neq I$, or $I neq (1)$. Formally, we have $c in mathbb C^n; c in V(I)$. That is, $eval_c(f) = 0$ for all $f in I$. But note that $eval_c(1) = 1$ for all $c$. Hence, we cannot have $1 in V(I)$. Therefore, $I neq (1)$.

Backward: $I neq (1) implies V(I) neq emptyset$:

Since $I neq (1)$, $I$ is contained in some maximal ideal $mathfrak m$. This ideal $mathfrak m$ has a point at which it vanishes, thus the ideal $I$, a subset of this $mathfrak m$ also vanishes on this point. Thus it cannot have empty vanishing set.

We know that $c in mathbb C^n$ such that $c in V(mathfrak m)$. Since $I subseteq mathfrak m$, $V(mathfrak m) subseteq mathfrak(I)$. Hence, $c in V(mathfrak m) subseteq mathfrak(I)$. Thus $c in mathfrak(I)$. Hence $I neq 0$.