I, beginning studying basic algebraic geometry by my own, and every book or notes Ive been reading began proving the following:
For an algebraic closed field $K$;
1.- $\emptyset$ and the whole affine space,$\mathbb{A}_{K}^{n}$ , are algebraic affine varieties.
2.- If $V(I)$ and $V(J)$ are algebraic affine varietes then $V(I) \cup V(J)$ is algebraic affine variety.
3.- If $\lbrace V(I_{i})\rbrace_{i \in I}$ is a family of algebraic affine variety, then $\bigcap_{i \in I} V(I_{i})$ is and algebraic affine variety.
For $1$, the sources Im reading prove the statement by mention that $\emptyset=V(1)$ and $\mathbb{A}_{K}^{n}=V(0)$, but I dont understand what $V(1)$ or $V(0)$ means?
For $2$, they prove $V(I) \cup V(J)=V(I \cap J)$ and I understand how they proved this double contention. But I dont understand why this proves $V(I) \cup V(J)$ is and algebraic variety.
For $3$, they prove $\bigcap_{i \in I} V(I_{i})= V(\sum_{i \in }I_{i})$ but as two what I dont understand is why this proves arbitrary intersection of algebraic affine varieties is indeed and algebraic variety.
Thanks!!!
1) Here, the author means $V((1))$ and $V((0))$, respectively where $(0), (1)$ are ideals.
2) & 3) Recall the definition of an algebraic variety. For 2) you show that $V(I) \cup V(J)$ is of the form $V(\cdot)$, namely $V(I \cap J)$. Varieties are defined to be sets of the form $V(\cdot)$. The same thing happens in 3).