I'm working on Hartshorne's Algebraic geometry exercise I.2.12. It says:
Let $Y$ be the image of the 2-uple embedding $\rho:\mathbb{P}^2\rightarrow\mathbb{P}^5$. If $C\subset Y$ is a closed curve (a curve is a variety of dimension 1), show that there exists a hypersurface (projective variety with dimension $n-1$ where $n=\dim \mathbb{P}^n$. This is equivalent to zero set of a single irreducible homogeneous polynomial) $V\subseteq\mathbb{P}^5$ such that $V\cap Y=C$
In here, $\rho$ is a map for $x=(x_0,x_1,x_2)$, $\rho(x)=(M_0(x),\cdots,M_5(x))=(x_0^2,x_1^2,x_2^2,x_0x_1,x_1x_2,x_0x_2)$
Since curve is a hypersurface in $\mathbb{P}^2$, there is a homogeneous irreducible polynomial $f\in k[x_0,x_1,x_2]$ ($k$ is a field) such that $C=\rho(\mathcal{Z}(f))$ (zeros of $f$).
Some website (here and question of here) says like this: Take $g=f^2\in k[M_0,\cdots M_5]$ in detail $f(x)^2=f(x_0,x_1,x_2)^2=g(M_0(x),\cdots M_5(x))$ and factorize $g=g_1\cdots g_r$ where each $g_i\in k[M_0,\cdots,M_5]$ is irreducible. (By here, each $g_i$ is homogeneous) Then there is one $g_i$, let's say $g_1=h$, such that $V=\mathcal{Z}(h)$ satisfies $V\cap Y=C$.
Showing $V\cap Y\subset C$ is clear: if $z\in V\cap Y$, there is $y\in\mathbb{P}^2$ such that $z=\rho(y)$ (by $z\in Y$) and $0=g_1(z)\cdots g_r(z)=g(z)$ (note $g_1(z)=h(z)=0$ by $z\in V$). So we get $0=g(z)=g(M_0(y),\cdots M_5(y))=f(y)^2$ which implies $f(y)=0$ and $y\in \mathcal{Z}(f)$. So $z=\rho(y)\in\rho(\mathcal{Z}(f))=C$.
My question starts from here: But $C\subset V\cap Y$ is unclear to me. If there is $z\in C$, we have for some $y\in \mathbb{P}^2$ such that $z=\rho(y)$ and $f(y)=0$. We get $g(z)=g(M_0(y),\cdots M_5(y))=f(y)^2=0$. But we cannot say $g_1(z)=0$ (because there may be other $g_i\neq g_1$ such that $g_i(z)=0$.) So we cannot say $z\in V$.
How can I pick specific irreducible $g_i$ (like in given pages) such that $\mathcal{Z}(g_i)= V$? Although this shows an answer, I want to check the claim in the above pages.
(Another thought: I think following is true: With the above setting, there are finite hypersurfaces $V_i\subset\mathbb{P}^5$ such that $(\bigcup_{i=1}^n V_i)\cap Y=C$ instead of the original problem.)
