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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.

enter image description here

(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.

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1 Answer 1

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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 $B$.

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  • $\begingroup$ 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 $\frac{\frac{k[x_1,..., x_n]}{A}}{B}\cong \frac{k[x_1,..., x_n]}{A+B}$? $\endgroup$
    – Nash
    Jun 20, 2019 at 0:47
  • $\begingroup$ 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 $I$. $\endgroup$
    – quantum
    Jun 23, 2019 at 19:21

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