# Commutative Algebra: homework

Last updated: Sat, 24 Nov 2018

(i) Prove the Nullstellansatz: $J \subsetneq k[x_1 ... x_n] \implies V(J) \neq \phi$ and $I(V(J)) = \text{ rad } J$ in an algebraically closed field $k[x_1 ... x_n]$.

To gain some intuition about this statement, consider the ideal $(x^2) \subset R[x, y]$; operating on it with V, we get y-axis as the variety; operating on the variety with I, we get the ideal $(x)$, which is the radical of $(x^2)$.

Proof: Let $f \in J \subset k[x_1 ... x_n]$ $(a_1 ... a_n) \in V(J)$ Introduce a new variable $y$, so that $J' = (J, fy - 1) \subset k[x_1 ... x_n, y]$ $(a_1 ... a_n, b) \in V(J')$ such that $bf(a_1 ... a_n) = 1$. Then, $V(J') = \phi \implies 1 = \sum g_i h_i + g_0 (fy - 1) \mid g_i \in k[x_1 ... x_n, y], h_i \in J$ Multiply both sides by $f^m$ to obtain: $f^m = \sum g_i(x_1 ... x_n, fy) h_i + g_0(x_1 ... x_n, fy) (fy - 1)$ This identity holds if $fy = 1$, and hence $f^m \in J$, as required $\Box$

(ii) Prove that a variety $X$ is irreducible if $I(X)$ is a prime ideal.

The definition of an irreducible variety is as follows. Let $X_1$ and $X_2$ be two varieties. $X = X_1 \cup X_2 \implies X = X_1 \vee X = X_2$

Proof: Let us proceed using proof by contradiction, and assume that $I$ is not prime. Let $f$ and $g$ be two prime ideals; i.e. $fg \in I \implies f, g \in A \backslash I$, for ring $A$. Now, define: $J_1 = (I, f), J_2 = (I, g)$

Then, $V(J_1) \subsetneq X, V(J_2) \subsetneq X$

which yields $X = V_1 \cup V_2$, a reducible, as required $\Box$