Decomposition of an algebraic variety into irreducible components

Question

I'm studying the Fulton's algebraic curves book and I have the following doubts in the end of the page 9:

I didn't understand why the following equations hold:

$$I\left(\bigcup_i V(F_i)\right)=\bigcap_i I(V(F_i))$$

$$\bigcap_i (F_i)=(F_1\cdots F_r)$$

Thanks in advance

Answer 1

Equation 1:

$$I\left(\bigcup_i V(F_i)\right)=\bigcap_i I(V(F_i))$$

Following the comments of the question, it's just a tautology:

$\subset$ part

$f\in I\left(\bigcup_i V(F_i)\right)\implies f(P)=0, \forall P\in \bigcup_i V(F_i)\implies f(P)=0,\forall P\in V(F_i)\ \text{for every i}\implies P\in I(V(F_i)),\ \text{for every i} \implies P\in \bigcap_i I(V(F_i))$

$\supset$ part

$f\in\bigcap_i I(V(F_i))\implies f\in I(V(F_i))\ \text{for every i}\implies f(P)=0\ \text{for}\ P\in V(F_i)\ \text{for every i}\implies f(P)=0\ \text{for}\ P\in \bigcup V(F_i)\implies f\in I\left(\bigcup_i V(F_i)\right)$

Equation 2:

$$\bigcap_i (F_i)=(F_1\cdots F_r)$$

$\subset$ part

For the case $r=2$: (the general case is similar)

Suppose $F_1, F_2$ are not associate:

$f\in (F_1)\cap (F_2)\implies f=g_1F_1=g_2F_2\implies F_1|g_2F_2\implies F_1|g_2$, the result follows.

(Note also that we use the fact that prime elements are irreducible and vice versa, because k[X,Y] is UFD)

$\supset$ part

Trivial