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I am having difficulty determining the dimension of a projective variety in general.

For example, I am confused about the dimension of the projective variety $X-Y=0$ in $\mathbb{P}^3$.

I was thinking that the dimension of this variety is $1$ because it defines a line. However, I am confused because if we also consider the line $Z=0$, then this is also of dimension $1$. But then the co-dimension of each variety is $2$, and $2+2=4>3$, implying that these varieties do not intersect (I'm not sure about this part - I know that when the co-dimensions add to less than or equal to $n$ that we have a non-empty intersection... is the converse true?). On the other hand, I know using common sense that these varieties intersect at the point $[1:1:0]$.

If someone could help me understand this example, I would really appreciate it. Please bear with me if I've made silly mistakes... I was only just introduced to these concepts. Thank you!

ET-phone-homology
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  • The dimension is two, because the dimension of the space is 3. In $\mathbb P^2$ a linear equation gives a line, in $\mathbb P^3$ gives a plane, etc. –  Apr 07 '15 at 04:53
  • The intersection of these linear spaces is the linear space $[t:-t:0:s]$ which is a line (as an intersection of plane !). –  Apr 07 '15 at 04:57
  • One last comment : the notion of dimension is local, if $U$ is open and $Y$ a variety (projective for example) then if $Y \cap U$ is not empty then $\dim(Y \cap U) = \ dim(Y)$ and you can focus on dimension in affine space. I hope it helps a bit ! –  Apr 07 '15 at 04:59
  • Thank you, your comments helped a lot! In $\mathbb{P}^n$, would the dimension of $X-Y=0$ be $n-1$ then? – ET-phone-homology Apr 07 '15 at 05:16
  • Yes. I put all is an answer ! –  Apr 07 '15 at 05:36
  • Last comment : you talk about $[1:1:0]$ but this point lives in $\mathbb P^2$, not $\mathbb P^3$. $\mathbb P^n$ coordinates are $[x_0: \dots : x_n]$. –  Apr 07 '15 at 06:27

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The dimension of a variety defined by a single equation $f=0$, called an hypersurface, highly depends of the dimension of the underlying space. In fact, such variety has dimension $n-1$ in $\mathbb P^n$. For taking your example again, $f = x-y = 0$ in $\mathbb P^n$, the linear subspace associated is $[t:-t:s_1:\dots:s_{n-1}]$. Assume for example $s_n = 1$ and you obtain $n-1$ parameters $t, s_1, \dots, s_{n-2}$ which shows that $\dim(Y) = n-1$. Essentially the last coordinate in $\mathbb P^1$ is just the point at infinity (in fact, in $\mathbb P^n$ the last coordinate is the hyperplane at infinity). For example, the line $[t:-t:0:s]$ is essentially the affine line $[t:-t:0:1]$ for $t \in k$ plus the point at infinity $[1:-1:0:0]$. So in comparaison in $\mathbb A^3$ you only added one point which clearly does not change the dimension. So you can focuse on dimension of affine variety, and here a "count of parameters" is enough for simple case.

  • I just want to make sure - is this the case for any $f$, or only linear ones? For example, does $XY-ZW=0$ in $\mathbb{P}^n$ also have dimension $n-1$ (for $n \geq 3$)? – ET-phone-homology Apr 07 '15 at 05:50
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    Sure ! The principle is very simple, informally. First, eliminate all the variables which don't appears in your polynomials. Here we can look $XY - ZW$ in $\mathbb P^3$. Now, look in an affine open, for example the affine open $W \neq 0$. Then we can assume $W = 1$ and we obtain the affine variety $Y$ defined by the equation $Z = XY$, and $Y$ is precisely the following subset of $\mathbb A^3$ $(x,y,xy)$, in particular of dimension $2 = 3 - 1$. Now, each variables you add will add one dimension to $Y$, which is given by $(x,y,xy, s_1, \dots, s_k)$ in $\mathbb P^{3+k} \cap {W \neq 0}$. –  Apr 07 '15 at 06:24
  • In fact, this variety is isomorphic to $\mathbb P^1 \times \mathbb P^1$ (Hint : look at the Segre Embedding) so this is another way of verify again that the zero set of $XY - ZW $ has dimension $2$ in $\mathbb P^3$ –  Apr 07 '15 at 06:26