# Prove the dimension of an affine algebraic variety X is equal to the dimension of its projective closure $\overline{X}$

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.

• 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
• Combo duplicate of this question and this question. 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
• 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. Does this resolve your question? – KReiser Nov 13 '20 at 1:10
• 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 9:45