# every irreducible component of $Y \cap H$ has dimension $r-1$

I am trying to show that "...every irreducible component of $$Y \cap H$$ has dimension $$r-1$$...", where the complete problem comes in the image, I have several doubts about the proof of this theorem

(I took this from http://sertoz.bilkent.edu.tr/courses/math591/2016/solutions.pdf , pag 1).

I know that this question is already posted on this site but my question is different.

(1) why $$\bar f=\bar f_1 \dots \bar f_s$$? Is this because $$B=k[x_1,..., x_n]/p$$ is a unique factorization domain?

(2) Why $$k[x_1,..., x_n]/[(f_i)+p]$$ is isomorphic to $$B/(\bar f_i)$$?

Thank you.

For 1), no $$B$$ may not be a UFD. In a Noetherian domain, every element can be written as a finite product of irreducible elements (irreducible does not mean prime, which is what will happen in a UFD).
For 2), What is your confusion? Going modulo $$(f_i)+p$$ is same as first going modulo $$p$$ (which gets you $$B$$) and then going modulo the image of $$f_i$$ which is $$\overline{f_i}$$ in $$B$$.
• If $A\subset k[x_1,..., x_n]$ is an ideal in $k[x_1,...,x_n]$ and $B$ is an ideal in $\frac{k[x_1,..., x_n]}{A}$, then $\frac{\frac{k[x_1,..., x_n]}{A}}{B}\cong \frac{k[x_1,..., x_n]}{A+B}$?
• Hint: Let $R$ be a commutative unitary ring and $I$ and ideal of $R$ and $\pi:R\rightarrow R/I$ be the natural surjection. Then any ideal $\bar J$ of $R/I$ is the $\pi$-image of an ideal $J$ of $R$ containing $I$. Jun 23, 2019 at 19:21