Identities about ideals

Question

If $A,B \subseteq K^n$ show the following:

• If $A \subseteq B$, then $I(B) \subseteq I(A)$.
• $I(A \cup B)=I(A) \cap I(B)$

Could you give me a hint, how the above identities could be proven?

EDIT: It is given that $I(X)=\{ f(x_1,x_2, \dots, x_n) \in K[x_1,x_2, \dots, x_n] \text{ such that } f(\overline{a})=0, \forall a \in X \}$

Answer 1

I think that $I(A)$ denotes the set $$ I(A)=\{f\in K[X_1,X_2,\dots,X_n]: f(a_1,a_2,\dots,a_n)=0 \text{ for all }(a_1,a_2,\dots,a_n)\in A\} $$

I'll simply use $f(a)$ for $f(a_1,a_2,\dots,a_n)$ where $a=(a_1,a_2,\dots,a_n)\in K^n$.

Suppose $A\subseteq B$ and take $f\in I(B)$; if $a\in A$, then $a\in B$, so $f(a)=0$; therefore $f\in I(A)$.

Suppose now $f\in I(A\cup B)$; if $a\in A$, then $f(a)=0$ and therefore $f\in I(A)$; if $b\in B$, then $f(b)=0$ and therefore $f\in I(A)\cap I(B)$.

I leave to you the proof that $I(A)\cap I(B)\subseteq I(A\cup B)$.

Answer 2

Let $f \in I(A \cup B)$ then $f(x) = 0$ for all $x \in A$. Same thing holds for all $x \in B$. So $f \in I(A) \cap I(B)$. Conversely if $f \in I(A) \cap I(B)$ then $f(x) = 0$ for all $x \in A$ and $f(x) = 0$ for all $x \in B$ and therefore $f(x) = 0$ for all $x \in A \cup B \Rightarrow f \in I(A \cup B)$.