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I'm trying to convince myself of the 4 following facts:

  1. If $X \subseteq Y \subseteq A^n_{k}$ then $I(Y) \subseteq I(X)$
  2. If $J \subseteq K[x_1,\ldots,x_n]$ is an ideal then $J \subseteq I(V(J))$
  3. If $X \subseteq A^n_{k}$ then $X \subseteq V(I(X))$
  4. $X=V(I(X))$ if and only if $X$ is an algebraic set

My attemps to prove them:
1) Let $f \in I(Y)$ then $f(p)=0$ for all $p\in Y$. So since $X \subseteq Y$ we have $f(p)=0$ for all $p \in X$, so $f \in I(X)$
2)
3)
4) -> If $X=V(I(X))$ then X= V(something) which is an algebraic set.
<- suppose an algerbaic set, i.e $X=V(J)$. Then by (2) $J \subseteq I(V(J))$.
so $I(V(J))=I(X)$.
so $X \subseteq V(I(X)) \subseteq V(J)$ by (3). So $X=V(I(X))$.

Is this right? I use part 2 and 3 for part 4, but cannot convince myself algebraicly why they hold.

1 Answers 1

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  1. Your argument is correct.

  2. Let $f\in J$. We need to prove that $f(p)=0$ for all $p\in V(J)$, so let $p\in V(J)$. By definition of $V(J)$, $g(p)=0$ for all $g\in J$, hence $f(p)=0$. Therefore, $f\in I(V(J))$.

  3. Let $p\in X$. We need to prove that $f(p)=0$ for all $f\in I(X)$, so to that end let $f\in I(X)$. By definition, this means that $f(q)=0$ for all $q\in X$, so in particular $f(p)=0$. Thus, $p\in V(I(X))$.

  4. $\implies$ is right. Your argument for $\impliedby$ is correct as well, but would be clearer if you add $V(J)=X$ at the end of your chain of inclusions, so you have "$X\subseteq V(I(X))\subseteq V(J)=X$".