Coordinate ring of an algebraic set.

Let $$K[x_1,x_2,....,x_n]$$ be a polynomial ring over an algebraically closed field $$K$$. Let $$V=V(I)$$ be an algebraic set in $$K^n$$ and $$I$$ is a radical ideal. We know by a theorem that there exists a hypersurface $$H_1=V(f_1)$$ such that dim$$(V\cap H_1)$$=dim$$(V)-1.$$

Then we continue intersecting with hypersurfaces to decrease the dimension of $$V$$ to $$0$$ i.e., there are hypersurfaces $$H_1, H_2,.., H_d$$ such that dim($$V\cap H_1\cap H_2\cap....\cap H_d) =0$$ where $$d=$$dim $$(V)$$ and $$H_i=V(f_i)$$. Now I want to see the behaviour of the corresponding coordinate rings. So, the coordinate ring of $$V$$ is $$\frac{K[x_1,x_2,....,x_n]}{I}$$ as $$I$$ is a radical ideal. First, I want to find the coordinate ring of $$V\cap H_1$$.

In a lecture video it is said that the coordinate ring of $$V\cap H_1=\frac{K[x_1,x_2,....,x_n]}{(I,f_1)}$$, but I think it should be $$\frac{K[x_1,x_2,....,x_n]}{\sqrt{(I,f_1)}}$$ because we don't know whether $$(I,f_1)$$ is a radical ideal or not. My ultimate aim is to understand the property that dim$$\frac{K[x_1,x_2,....,x_n]}{(I,f_1,f_2,..,f_d)}=0$$. So, my question is whether $$(I, f_1)$$ is a radical ideal or not?

Thank you

It is possible to choose $$f_1$$ such that $$(I,f_1)$$ is radical and $$V(I)\cap V(f_1)$$ has dimension one less than $$V(I),$$ with $$\dim(\emptyset)=-1.$$ A general affine function will do the job. This is a variant of Bertini's theorem. Let $$X$$ be the projectivization of $$V(I)\subset \mathbb A^n_K,$$ so $$X$$ is a reduced closed subscheme of $$\mathbb P^n.$$ Then $$X\cap H$$ is reduced for a general hyperplane $$H$$ in $$\mathbb P^n.$$ See for example Corollary 2 of Cumino, Caterina; Greco, Silvio; Manaresi, Mirella, An axiomatic approach to the second theorem of Bertini, J. Algebra 98, 171-182 (1986). ZBL0613.14006. Reducedness is affine-local, so $$X\cap H\cap \mathbb A^n$$ will be reduced, and $$H\cap \mathbb A^n$$ is an affine hyperplane for general $$H.$$
No, $$(I, f)$$ is not necessarily a radical ideal. Take, for instance, the radical ideal $$I = (xy)\subseteq \Bbb C[x, y]$$, and let $$f_1 = x-y$$. Then $$(I, f_1) = (x^2, x-y)$$, which is not radical, as it contains $$x^2$$ but not $$x$$.
That being said, it would surprise me greatly if there was no $$f_1$$ such that $$(I, f_1)$$ is proper and radical.
• Ok, so here we are choosing $f_1$ such $V\cap H_1$ has dim $d-1$ and $(I, f_1)$ is proper and radical. Mar 21, 2019 at 10:42