We know the dimension of a affine algebraic variety $X\subseteq$ $\mathbb{A}^{n}_k$ (k a field, and $X$ does not have to be irreducible) is the biggest integer r such that there exists a strictly increasing chain $Z_{0}\subset Z_{1}\subset ... \subset Z_{r}\subset X$ of affine algebraic varieties in X. On the other side, $\overline{X}$ $\subseteq$ $\mathbb{P}^{n}_k$ is the smallest algebraic subset in $\mathbb{P}^{n}_k$ (k a field) such that $X\subseteq$ $\overline{X}$. I tried proving that the projective closure preserve the inclusions, i.e., if $Z\subseteq Y$, for $Z, Y\subseteq X$, then $\overline{Z}\subseteq \overline{Y}$, for $\overline{Z},\overline{Y}\subseteq\overline{X}$, but I don't know how to proceed or if this will help me with the initial problem.

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    You have a chain of ideals $I_0 \supset I_1 \supset \cdots \supset I_r \supset I(X)$ that you can homogenize. Basically what you want to check is that homogenizing preserves containment of ideals. – Tabes Bridges Nov 12 '20 at 19:46
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    Combo duplicate of [this question](https://math.stackexchange.com/questions/1789033/dimension-of-irreducible-affine-variety-is-same-as-any-open-subset) and [this question](https://math.stackexchange.com/questions/140066/krull-dimension-of-a-scheme/140078). The first says that all the intersection of all standard affine open patches with the closure have the same dimension as $X$, and then the second says that dimension can be checked on an open cover. – KReiser Nov 12 '20 at 19:58
  • I understand the second question but I don't know what do you mean by "the intersection of all standard affine open patches with the closure"? I cant deduce that on the first question you add – Laura Nov 12 '20 at 21:15
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    You need to use @ to ping other users if you're attempting to reply to them (unless you're commenting on their post) otherwise they won't know that you've responded. I've found a more suitable duplicate target [here](https://math.stackexchange.com/questions/3507784/open-nonempty-subset-of-irreducible-variety-over-k-has-the-same-krull-dimensio). Does this resolve your question? – KReiser Nov 13 '20 at 01:10
  • @KReiser One thing I just realized is that I'm asking for an affine variety but does not have to be irreducible, and the results you attached are for irreducible ones, so what can I do in this case? – Laura Nov 13 '20 at 09:40
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    First, please include this information in your post next time. Next, you can go component-by-component: if $X=X_1\cup\cdots\cup X_n$ is a decomposition in to irreducible components, $\overline{X}=\overline{X_1}\cup\cdots\cup\overline{X_n}$ is a decomposition in to irreducible components. Applying the result, $\dim X_i=\dim \overline{X_i}$ and as dimension is the maximum of the dimension of the irreducible components, we're done. – KReiser Nov 13 '20 at 09:45
  • @JoséCarlosSantos Why did you reopen this? – KReiser Nov 13 '20 at 11:06
  • @KReiser Last question..., if the decomposition in to irreducible components of $X$ is minimal, how do you know that the decomposition in to irreducible components of $\overline{X}$ is minimal too? – Laura Nov 13 '20 at 14:18
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/116201/discussion-between-laura-and-kreiser). – Laura Nov 13 '20 at 17:33

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