Define $R:=\mathbb{C}[x_0,x_1,\cdots]/(x_0x_1,x_1x_2,\cdots,x_{2n}x_{2n+1},\cdots)$ .
R is a quotient ring of the ring of polynomials over complex field in inifitete variable.
I found this ring in Example 167 in the book "ExamPles of Commutative Rings" by Harry C. Hutchins.
Is Spec$R$ noetherian topological space? If not, give me a strictly decreasing chain of closed subsets in Spec$R$
The only thing I know about this ring is that the minimal prime ideal of R are those of the form $P=(x_{0+d_1},x_{2+d_2},\cdots,x_{2n+d_{2n}},\cdots)$ where $d_i$ is either $0$ or $1$.
Expanding on Cranium Clamp's comment: If you mod out all the even-indexed variables, you obtain $$R/(x_{2k} | k \in \mathbb N_0) \cong \mathbb C[x_1, x_3, x_5, \dotsc],$$ because the ideal $(x_0x_1, x_1x_2, x_2x_3, \dotsc)$ is contained in the ideal $(x_0, x_2, x_4, \dotsc)$ in $\mathbb C[x_0, x_1, x_2, \dotsc]$. This means that $\operatorname{Spec} R$ contains $Z = \operatorname{Spec} \mathbb C[x_1, x_3, \dotsc]$ as a closed subset. As the latter is not noetherian, $\operatorname{Spec}R$ cannot be noetherian. You obtain an infinitely decreasing sequence of closed subsets by choosing it inside of $Z$.