111.16 Hilbert Nullstellensatz
Exercise 111.16.1. A silly argument using the complex numbers! Let ${\mathbf C}$ be the complex number field. Let $V$ be a vector space over ${\mathbf C}$. The spectrum of a linear operator $T : V \to V$ is the set of complex numbers $\lambda \in {\mathbf C}$ such that the operator $T - \lambda \text{id}_ V$ is not invertible.
Show that $\mathbf{C}(X)$ has uncountable dimension over ${\mathbf C}$.
Show that any linear operator on $V$ has a nonempty spectrum if the dimension of $V$ is finite or countable.
Show that if a finitely generated ${\mathbf C}$-algebra $R$ is a field, then the map ${\mathbf C}\to R$ is an isomorphism.
Show that any maximal ideal ${\mathfrak m}$ of ${\mathbf C}[x_1, \ldots , x_ n]$ is of the form $(x_1-\alpha _1, \ldots , x_ n-\alpha _ n)$ for some $\alpha _ i \in {\mathbf C}$.
Exercise 111.16.3. Let $k$ be a field. Please use Remark 111.16.2.
Let $R$ be a $k$-algebra. Suppose that $\dim _ k R < \infty $ and that $R$ is a domain. Show that $R$ is a field.
Suppose that $R$ is a finitely generated $k$-algebra, and $f\in R$ not nilpotent. Show that there exists a maximal ideal ${\mathfrak m} \subset R$ with $f\not\in {\mathfrak m}$.
Show by an example that this statement fails when $R$ is not of finite type over a field.
Show that any radical ideal $I \subset {\mathbf C}[x_1, \ldots , x_ n]$ is the intersection of the maximal ideals containing it.
Exercise 111.16.5. Let $A = {\mathbf C}[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}]$. Let $I$ be the ideal of $A$ generated by the entries of the matrix $XY$, with
\[ X = \left( \begin{matrix} x_{11}
& x_{12}
\\ x_{21}
& x_{22}
\end{matrix} \right) \quad \text{and}\quad Y = \left( \begin{matrix} y_{11}
& y_{12}
\\ y_{21}
& y_{22}
\end{matrix} \right). \]
Find the irreducible components of the closed subset $V(I)$ of $\mathop{\mathrm{Spec}}(A)$. (I mean describe them and give equations for each of them. You do not have to prove that the equations you write down define prime ideals.) Hints:
You may use the Hilbert Nullstellensatz, and it suffices to find irreducible locally closed subsets which cover the set of closed points of $V(I)$.
There are two easy components.
An image of an irreducible set under a continuous map is irreducible.
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