Some very basic problems in understanding the definition of algebraic variety

Question

I just started learning myself some basic algebraic geometry and I have some trouble doing these rather elementary exercises. I am basically misunderstanding something fundamental so it's causing me trouble to progress further.

Some definitions: $X\subset k^n$ is an affine (algebraic) variety if $X=\mathbb{V}(I)$ for some ideal $I\subset R,$ where $\begin{aligned}\mathbb{V}(I)=\{a\in k^n\mid f(a)=0\text{ for all }f\in I\}\end{aligned}$.

Exercises:

$(1)\:I\subset J\Rightarrow\mathbb{V}(I)\supset\mathbb{V}(J).\:($“The more equations you impose, the smaller the solution set”.)

$(2)\quad\mathbb{V}(I)\cup\mathbb{V}(J)=\mathbb{V}(I\cdot J)=\mathbb{V}(I\cap J).$

(3) $\mathbb{V}(I)\cap\mathbb{V}(J)=\mathbb{V}(I+J).$ (Note: $\langle I\cup J\rangle=I+J.)$

(4) $\mathbb{V}(I),\mathbb{V}(J)$ are disjoint if and only if $I,J$ are relatively prime (i.e. $I+J=\langle1\rangle)$

My thought processes:

1. I don't think I understand why it's $\mathbb{V}(I)\supset\mathbb{V}(J),$ rather than $\mathbb{V}(I)\subset\mathbb{V}(J).$

2. The union of these two sets represents the set of points where at least one polynomial in either $I$ or $J$ vanishes. I don't see how's it equal to the product of two ideals and the intersection of them.

3. Not sure why's that. If I had to guess, I'd likely say that $\mathbb{V}(I)\cap\mathbb{V}(J)=\mathbb{V}(I\cdot J),$ but that's probably wrong. 3

4. If $\mathbb{V}(I)$ and $\mathbb{V}(J)$ are disjoint, it means there are no common points between them. Common points between $\mathbb{V}(I)$ and $\mathbb{V}(J)$ correspond to solutions of polynomials in both $I$ and $J$. If $\mathbb{V}(I)$ and $\mathbb{V}(J)$ are disjoint, there are no common solutions, which implies that $I + J$ contains only constant polynomials (since there are no common solutions, $I + J$ cannot generate non-constant polynomials). Therefore, $I + J = \langle 1 \rangle$, as any non-constant polynomial would introduce common solutions.

Can someone help me understand this better or at least guide me in the right direction? Thanks.

Answer 1

Part 4 requires $k$ to be algebraically closed. Otherwise, you can take $k=\mathbb{R}$ and the ideal $(x_1^2+1)$.

You seem to be confusing "for all" and "for some" in a lot of places...

1. If $I\subseteq J$, then something that happens for all $f\in J$ will also happen for all $f\in I$, but not necessarily the other way around. So if a point $a\in k^n$ satisfies that $f(a)=0$ for all $f\in J$, then it will also satisfy that $f(a)=0$ for all $f\in I$ (because everything that is in $I$ is also in $J$). That gives $\mathbb{V}(J)\subseteq \mathbb{V}(I)$, not the other way around.

2. No, the union $\mathbb{V}(I)\cup\mathbb{V}(J)$ is not the set of points where at least one polynomial in $I$ or at least one polynomial in $J$ vanishes. It is the set of points for which either all polynomials in $I$ vanish, or all polynomials in $J$ vanish. That's because a point $a\in k^n$ lies in the union if and only if it lies in either $\mathbb{V}(I)$ or in $\mathbb{V}(J)$. Now, since $IJ\subseteq I\cap J\subseteq I$, then from (i) you know that $\mathbb{V}(I)\subseteq \mathbb{V}(I\cap J)\subseteq \mathbb{V}(IJ)$. And likewise, since $IJ\subseteq I\cap J\subseteq J$, you also know $\mathbb{V}(J)\subseteq \mathbb{V}(I\cap J)$. Since $\mathbb{V}(I)\subseteq \mathbb{V}(I\cap J)$ and $\mathbb{V}(J)\subseteq \mathbb{V}(I\cap J)$, you also know that $\mathbb{V}(I)\cup\mathbb{V}(J)\subseteq \mathbb{V}(I\cap J)$. So to prove the equality, you just need to show that $$\mathbb{V}(IJ)\subseteq \mathbb{V}(I\cap J)\subseteq \mathbb{V}(I)\cup\mathbb{V}(J).$$ To show that $\mathbb{V}(I\cap J)\subseteq \mathbb{V}(I)\cup\mathbb{V}(J)$, let $a\in k^n$ be such that every element of $I\cap J$ vanishes at $a$. If every element of $I$ vanishes at $a$, then $a\in\mathbb{V}(I)$ and we are done. If this is not the case, let $f\in I$ be such that $f(a)\neq 0$. Now, for every $g\in J$ we have $fg\in I\cap J$, hence $(fg)(a) = f(a)g(a)=0$. But $f(a)\neq 0$, so for this product to be $0$ we must have $g(a)=0$. Thus, for all $g\in J$ we have $g(a)=0$, hence $a\in\mathbb{V}(J)$. Thus, if $a\in\mathbb{V}(I\cap J)$, then either $a\in\mathbb{V}(I)$ or $a\in\mathbb{V}(J)$. This proves $\mathbb{V}(I\cap J)\subseteq \mathbb{V}(I)\cup\mathbb{V}(J)$.
   
   To show that $\mathbb{V}(IJ)\subseteq \mathbb{V}(I\cap J)$, let $a\in\mathbb{V}(IJ)$. If there is an $f\in I\cap J$ such that $f(a)\neq 0$, then $f^2\in IJ$, and $f^2(a) = f(a)f(a)\neq 0$, so $a\notin\mathbb{V}(IJ)$, a contradiction. Therefore, $f(a)=0$ for all $f\in I\cap J$, so $a\in\mathbb{V}(I\cap J)$, as desired.

3. Since $I\subseteq I+J$, then $\mathbb{V}(I+J)\subseteq\mathbb{V}(I)$ by (1). Likewise, $\mathbb{V}(I+J)\subseteq \mathbb{V}(J)$. Since $\mathbb{V}(I+J)$ is contained in both $\mathbb{V}(I)$ and $\mathbb{V}(J)$, it is contained in their intersection; that gives $\mathbb{V}(I+J)\subseteq \mathbb{V}(I)\cap\mathbb{V}(J)$. Conversely, if $a\in\mathbb{V}(I)\cap\mathbb{V}(J)$, then every $f\in I$ and every $g\in J$ satisfies $f(a)=g(a)=0$. Let $h\in I+J$. Then we can write $h=f+g$ with $f\in I$ and $g\in J$; so $h(a) = (f+g)(a) = f(a)+g(a) = 0+0 = 0$. Thus, every $h\in I+J$ satisfies $h(a)=0$, hence $a\in\mathbb{V}(I+J)$, as claimed.

4. Again, you seem to be confusing a "for all" statement with an "there exists" statement. If $\mathbb{V}(I)$ and $\mathbb{V}(J)$ are disjoint, then there is no point which is a zero for both all polynomials in $I$ and all polynomials in $J$. That means that given a point $a\in k^n$, you can find at least one $f\in I$ and at least one $g\in J$ such that $f(a)\neq 0$ and $g(a)\neq 0$. It does not mean that $a$ is not the root of any polynomial in $I$ or in $J$.
   
   To prove this we can use (3): if $I+J=(1)$, then $\mathbb{V}(I+J) = \mathbb{V}(1) = \varnothing$, because no point satisfies $1(a) =0$. Therefore, $\mathbb{V}(I)\cap\mathbb{V}(J) = \mathbb{V}(I+J)=\varnothing$. Conversely, if $\mathbb{V}(I)\cap\mathbb{V}(J)=\varnothing$, we want to prove that $I+J=(1)$. We argue by contrapositive: if $I+J\neq (1)$, then there is a maximal ideal $M$ such that $I+J\subseteq M$, and therefore $\mathbb{V}(M)\subseteq \mathbb{V}(I+J)=\mathbb{V}(I)\cap\mathbb{V}(J)$. By Hilbert's Nullstellensatz, $M$ is of the form $(x_1-a_1,\ldots,x_n-a_n)$ for some $(a_1,\ldots,a_n)\in k^n$. But then $(a_1,\ldots,a_n)\in\mathbb{V}(M)$, so $\mathbb{V}(I)\cap\mathbb{V}(J)\neq\varnothing$.

Answer 2

To get a good initial idea, play with unbelievably simple examples.

For example, maybe I'll look at $k = \mathbb{R}$, choose polynomials in $R = \mathbb{R}[X, Y]$, and look at resulting ideals. (Later it will turn out that there are complications from not using an algebraically closed ring $k$, in which case you might look at $k = \mathbb{C}$. But $\mathbb{C}^2$ is hard to visualize while $\mathbb{R}^2$ is familiar).

For example, let $I = \langle X \rangle$, $J = \langle Y \rangle$, $A = \langle X, Y \rangle$, and $B = \langle XY \rangle$ be ideals. What is $V(I)$, the zero set of $I$? It's the $y$-axis, consisting of points $\{ (x, y) \in \mathbb{R}^2 : x = 0 \}$. Similarly $V(J)$ is the $x$-axis.

What about $V(A)$ and $V(B)$? Any point $(x, y) \in V(A)$ must have $x = 0$, as $X \in A$; and it must have $y = 0$, as $Y \in A$. Thus $V(A) = \{ (0, 0) \}$ is just the origin. Similar reasoning shows that any points $(x, y) \in V(B)$ must either have $x = 0$ or $y = 0$, and so $V(B)$ is the union of the $x$-axis and the $y$-axis.

Finally, observe that $I \cup J = A$ and $I \cap J = B$. We can use the examples about to reason about the first three properties you mention (the fourth involves relatively primeness and I haven't touched on).

1. In my notation, note that $I \subset A$. $V(I)$ is the $y$-axis and $V(A)$ is the origin, a single point. We see $V(A) \subset V(I)$, but not the other way around. That tagline next to this property is a good reason why: for a point $(x, y)$ to be in $V(A)$, it must satisfy $f(x, y) = 0 \forall f \in A$. As $A$ is larger than $I$, this is a more restrictive property than satisfying $f(x, y) = 0 \forall f \in I$.

2. In my notation, note that $I\cdot J = B$. And $V(I)$ is the $y$-axis, $V(J)$ is the $x$-axis, and $V(B)$ is the union of the two axes. Reasoning about $V(B)$ is almost exactly the proof that $V(I \cdot J) = V(I) \cup V(J)$ in general.

3. In my notation, $I \cup J = A$. The intersection of $V(I)$ [the $y$-axis] and $V(J)$ [the $x$-axis] is the origin [$V(A)$]. Looking back on my reasoning showing that $V(A)$ is a single point, I noted that $(x, y) \in V(A)$ needs to satisfy both the requirements for $V(I)$ and for $V(J)$.

Other good examples to look at to build intuition are circles $X^2 + Y^2 - 1$ and lines $aX + bX + c$.