Proposition

Let $X = \mathbb{C}^n$ and $A = \mathbb{C}[x_1, \cdots, x_n]$.

Maximal ideals of $A$ are in one-to-one correspondence with points $P \in X$. That is

Proof

Unfortunately, I don’t know how to prove this, but I found this proposition in P.5 of Undergraduate Commutative Algebra. This is a special case of the weak Nullstellensatz.

Example

Although I don’t know how to prove this, this proposition is believable.

Let $P = (1, 2, 3)$. Then $m_P = (x_1 - 1, x_2 - 2, x_3 - 3)$. In other words, $m_P$ is the ideal of $A$ generated by three polynomials $x_1 - 1, x_2 - 2, x_3 - 3$. Then $m_P$ is a maximal ideal.

Note: $m_P$ is NOT generated by $(x_1 - 1)(x_2 - 2)(x_3 - 3)$. It is generated by the three polynomials.

On the other hand, let $m \subset A$ be a maximal ideal. I don’t know how to prove it, but it is believable that $m$ is generated by three polynomials $x_1 - a_1, x_2 - a_2, x_3 - a_3$ because:

• $\mathbb{C}$ is algebraically closed. If $m$ is generated by a polynomial of degree 2 such as $x^2 + 1$, then the ideal $(x - i, x + i)$ will properly contain $m$. Since $m$ is a maximal ideal, $m$ cannot be generated by a polynomial of degree 2 or above.
• The ideal $(x_1 - a_1, x_2 - a_2)$ is property contained in $(x_1 - a_1, x_2 - a_2, x_3 - a_3)$. Therefore, it sort of makes sense that $m$ is generated by three polynomials each of which contains exactly one variable.

Why is this important?

This shows some connection between algebraic objects and geometric objects.