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Given sub-variety $X\subset \mathbb{A}^{2k}$ of dimension $k-1$, how can I find a sub-variety $Y\subset \mathbb{A}^{2k}$ of the same dimension which is disjoint to $X$?

Perhaps I should mention I read Kempf's 'algebraic varieties' up to chapter 5.

pumpam
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1 Answers1

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I don't have Kepf's book, but if by an algebraic variety you mean a Zariski closed subset of the affine space (over an algebraically closed field), then here is one way to do it.

Let $X=\mathrm{Spec}(k[x_1,\cdots,x_{2k}]/I)$. If $\dim(X)=k-1$, then you can find a lot of closed subsets of dimension 1 in $\mathbb{A}^{k+1}=\mathrm{Spec}(k[x_1,\cdots,x_{k+1}])$ whose intersection with $X$ has dimension at most $0$ (in fact most irreducible $1$ dimensional varieties will be like that). Pick one and call it $Y$. This will be a closed subset defined by polynomials $f_i(x_1,\cdots,x_{k+1})$. Note that $Y\times \mathbb{A}^{k-1}$ is a variety of dimension $k$ in $\mathbb{A}^{2k}$ whose intersection with $X$ has dimension at most $k-1$.

Let $g_i$'s be the generators of $I$ and denote by $g$ their product. Now intersect $Y\times \mathbb{A}^{k-1}$ with the codimension 1 subset $\mathrm{Spec}(k[X]/(x_{k+2}g-1))$. The intersection has dimension $k-1$ and does not intersect $X$.

adrido
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  • Thanks abdrido for your answer. I managed to follow you up to the last paragraph. Could you please explain what you mean by $x_{k+2g}$ ? – pumpam Jan 28 '14 at 12:57
  • @pumpam: it's product of the indeterminate $x_{k+2}$ and $g$ (product of generators of $I$). I'm just adding an ``inequation'' $g\neq 0$, to ensure that there is no intersection. – adrido Jan 28 '14 at 17:25
  • I see...for some reason I read it $x_{k+(2g)}$ which is silly. But shouldn't it be $k[\mathbb{A}^{2k}]$? – pumpam Jan 28 '14 at 20:42
  • that works too, but i wanted to emphasize that you can get a subvariety inside $A^{k+1}$ and still have lots of free variables to impose the inequation $g\neq 0$ (in $\mathbb{A}^{2k}$). – adrido Jan 28 '14 at 23:34