Equivalence of Classical Nullstellensatz to “Affine schemes have points”

algebraic-geometrycommutative-algebraring-theory

For $k$ an algebraically closed field and $I$ a proper ideal in the polynomial ring $k[X_1, \dots, X_n]$, the set $V(I)$ (of $n$-tuples $\vec x \in k^n$ such that all polynomials in $I$ vanish when evaluated on these $\vec x$) is an inhabited set.

we remark that an element of $V(I)$ is just a $k$-algebra homomorphism of the form
$k[X_1, \dots X_n]/I \rightarrow k$.

I believe the map we are discussing is to consider each element $\vec x \in V(I)$ as the evaluation homomorphism , $\phi_{\vec x}: K[X_1, X_2, \dots X_n]/I \rightarrow k$ is the evaluation map which maps $\phi(p) = p(\vec x)$.

I don't understand why we need to quotient by the ideal $I$. Even without the quotient, it continues to be a homomorphism? We have that:

$$
\forall p, q \in K[X_1, \dots, X_n], \\
(p + q)(\vec x) = p(\vec x) + q(\vec x) \\
(p \cdot q)(\vec x) = p(\vec x) \cdot q(\vec x) \\
$$

  1. So, why do we bother quotienting with $I$? What am I missing here?

They go on to say:

Dually this is a morphism of affine schemes (ring spectra) of the form
$\operatorname{Spec}(k) \rightarrow \operatorname{Spec}(k[X_1, \dots X_n] / I)$. Moreover since $\operatorname{Spec}(k)$ is the terminal object in this context, such a map is the same as a "point", a global element of $\operatorname{Spec}(k[X_1, \dots X_n] / I)$. Hence in this form the Nullstellensatz simply says that (for $k$ algebraically closed) affine schemes have points

I am quite lost at this stage.

  1. I understand that $Spec(k)$ contains only the zero ideal $(0)$ since $k$ is a field, and hence the morphism is as good as singling out a single point. However, in what way is this a "terminal object"? In what category even are we discussing the above?

  2. How is this equivalent to the (only) Nullstellensatz that I know, which states:

Nullstellensatz, statement 1: maximal ideals of $\mathbb C[X_1, \dots X_n]$ are in bijection with points in $\mathbb C^n$?

Nullstellensatz, statement 2: All maximal ideals of $\mathbb C[X_1, \dots X_n]$ are of the form $(x – C_1, x – C_2, \dots x- C_n)$ for $c_i \in \mathbb C$

I want to understand this form of the Nullstellensatz since it is the one that is used in "Yuri Manin, Introduction into theory of
schemes". The textbook contains this as an exercise. I'm unfortunately unable to make sense of this.

Best Answer

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 $V(\mathfrak m) = \{ c\star \}$ for some $c\star \in \mathbb C^n$. For contradiction, assume not.
  3. Then consider the ideal $J = (X_1 - c\star_1, \dots, X_n - c\star_n)$.
  4. $V(J) = \{ c\star_n \}$. Clearly, $c\star \in V(J)$ since the polynomials $X_i - c\star_i$ vanish at $c\star$. 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 $d_i - c\star_i = 0$. This implies $d = c\star$.
  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 $I \neq 0$.

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