# Curves on the projective plane

I have two little questions, I'm learning this, and I'm not accustomed yet )=. The questions are so simple. First define on $$R^3 - \left\{ {\left( {0,0,0} \right)} \right\}$$ the topology given by the relation $$\left( {x,y,z} \right) \sim \left( {tx,ty,tz} \right)$$ call this new space $$P^2 _R$$ Given an homogeneous polynomial $F(x,y,z)$ over $R$. Then define $$C = \left\{ {\left[ {x,y,z} \right] \in P^2 _R /F\left( {x,y,z} \right) = 0} \right\}$$ I want to prove that $C$ is closed. I have no idea, clearly if the set was the set $$D = \left\{ {\left( {x,y,z} \right) \in R^3 /F\left( {x,y,z} \right) = 0} \right\}$$ being the preimage of zero, would be closed, by intuition I guess you should use that and something else but I do not know. T_T Excuse me for these questions, but still can not get used to this. And also prove that if $F(x,y,z ) = ax+by+cz$ then two curves of this kind of function, always intersect, and only in one point. Sorry for ask this simple question. But I want to see some examples. Thanks

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Dear Daniel, Thinking about $D$ is a good start. Now, what does it mean for a subset of a quotient, equipped with the quotient topology, to be closed? Regards, – Matt E Sep 13 '11 at 0:05
Dear Matt, I thought they removed all salutations, like Dear, from MSE? Hmm. I tried it here, and we see the result. – mixedmath Sep 13 '11 at 0:19

Going up to $\mathbf R^3 \setminus \{0\}$ and $\mathbf R^3$ is the right thing to do. I presume that your definition of a quotient map is a a map $f\colon X \to Y$ of topological spaces such that $U \subset Y$ is open if and only if $f^{-1}(U)$ is open. But it is equivalent to require that $A \subset Y$ be closed if and only if $f^{-1}(A)$ is closed. (Check this!)
So I would try to think of the relationship between $C$ and $D \setminus \{0\} \subset \mathbf R^3 \setminus \{0\}$.
This advice also applies to your particular example. In $\mathbf R^3$, for $a, b, c$ not all zero, the zero set of $ax + by + cz$ is a two-dimensional vector subspace. If you have two distinct subspaces like this, what is their intersection? In particular, what dimension will the intersection have? This is just a problem in linear algebra.
Oh right, Thanks Matt, will be closed iff $$p^{ - 1} \left( A \right)$$ is closed, in this case, $$p^{ - 1} \left( C \right) = D$$ so it´s done O= – Daniel Sep 13 '11 at 0:22
@Daniel Careful! It's $D$ less the origin. – Dylan Moreland Sep 13 '11 at 0:28