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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

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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 $H.$

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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.

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  • $\begingroup$ 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. $\endgroup$
    – uuuuuuuuuu
    Mar 21, 2019 at 10:42

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