Formula of intersection of homogeneous ideal

Question

I found that I misread the concept about homogeneous ideal in the projective space in the Hartshorne's Algebraic Geometry (this book).

For any subset $Y$ in the projective space $\mathbb{P}^n$,the definition of homogeneous ideal of $Y$, denoted $I(Y)$ is the ideal generated by $$\{f\in k[x_0,\cdots x_n]:f\text{ is homogeneous and }f(P)=0 \text{ for all }P\in Y\}$$ where $k$ is an algebraically closed field.

I thought $$I(Y)=\{f\in k[x_0,\cdots x_n]:f(P)=0\text{ for all }P\in Y\}.$$ The exercise (chapter 1 2.3.(c)) asks to show $$I(Y_1\cup Y_2)=I(Y_1)\cap I(Y_2)$$ for any two subsets $Y_1,Y_2\subseteq \mathbb{P}^n$. My wrong definition satisfies this equation for very simple reason. With original definition, I could show $I(Y_1\cup Y_2)\subseteq I(Y_1)\cap I(Y_2)$. However I am stuck at showing $I(Y_1\cup Y_2)\supseteq I(Y_1)\cap I(Y_2)$. Following is what I have tried.

If $f\in I(Y_1)\cap I(Y_2)$, $f=\sum_i a_if_i=\sum_j b_j g_j$ where $a_i,f_i,b_j,g_j\in k[x_0,\cdots x_n]$, $f_i,g_j$'s are homogeneous and $f_i(P)=0$ for all $P\in Y_1$ and $g_j(Q)=0$ for all $Q\in Y_2$. (This comes from the definition of $f\in I(Y_1)$ and $f\in I(Y_2)$.

If we show $f_i(T)=g_i(T)=0$ for all $T\in Y_1\cup Y_2$, then $f\in I(Y_1\cup Y_2)$. But we don't know that.

Two things I found are

1. $f^2=\sum_{ij}a_ib_jf_ig_j$ and $f_ig_j(T)=0$ for all $T\in Y_1\cup Y_2$ and $f_ig_j$'s are homogeneous. Thus $f^2\in I(Y_1\cup Y_2)$.
2. If $f$ is homogeneous, $f(T)=0$ for all $T\in Y_1\cup Y_2$ (it also holds when $f$ is not homogeneous). Since the element $f$ is generated by a homogeneous polynomial $f$, $f\in I(Y_1\cup Y_2)$.

Can anyone give a hint to $I(Y_1\cup Y_2)\supseteq I(Y_1)\cap I(Y_2)$?

Answer 1

Thanks to user6565190, I found a solution, I give an answer.

Write $f=f_0+\cdots+f_m$ where $f_i$ is the homogeneous polynomial with degree $i$. Then above gives $$f=f_0+\cdots+f_m=\sum_ia_if_i=\sum_j b_jg_j\cdots (*)$$. For $T\in Y_1$ define a polynomial $T(\lambda)=f_0(\lambda T)+\cdots f_m(\lambda T)=f_0(T)+\cdots+\lambda^mf_m(T)$. It is $0$ by the equation $(*)$ for every $\lambda\in k\setminus\{0\}$. Since $k$ is algebraically closed, it is infinite. Thus every coefficient of $T(\lambda)$, which is $f_n(T)$, should be zero ($0\leq n\leq m$). So $f_n\in I(Y_1)$. Similarly $f_n\in I(Y_2)$. By 2 in my original post, $f_n\in I(Y_1\cup Y_2)$ Thus $f=\sum_nf_n\in I(Y_1\cup Y_2)$.