### Examples- II

The set of irritating examples continues:

1. $V(I\cap J)=V(I.J)=V(I)\cup V(J)$: let $I$ be the ideal generated by the polynomial $x+y=0$ and $J$ be the polynomial generated by $x-y=0$. Then $I\cap J$ consists of the polynomials that are present in both ideals. As $I$ and $J$ are both prime ideals, their intersection is exactly the product of the two ideals. When we take a product of two ideals, their set of common zeroes is the union of the set of zeroes of the individual ideals. Hence, we get $V(I\cap J)=V(I)\cup V(J)$.

Why is the intersection of two prime ideals equal to their product? It is easy to see that the product of the two ideals would be contained within the intersection. But what if the intersection is bigger than the product?

2. Explicitly write down a morphism between two varieties, that leads to a morphism between their coordinate rings: Consider the map $t\to (t,t^2)$, which is a morphism between the affine varieties $\Bbb{R}$ and $V(y-x^2)\subset \Bbb{R}^2$. We should not construct a morphism between the coordinate rings $\Bbb{R}[x,y]/(y-x^2)$ and $\Bbb{R}[x]/(0)=\Bbb{R}[x]$. How do we go about doing that?
Consider any polynomial in $\Bbb{R}[x,y]/(y-x^2)$; say something of the form $x^2+y^2$. We can now replace $x$ by $t$ and $y$ by $t^2$. That is how we get a morphism from $\Bbb{R}[x,y]/(y-x^2)$ to $\Bbb{R}[x]$.

Now we shall start with a morphism between coordinate varieties. Consider the morphism $\Bbb{R}[x,y]/(y-x^2) \to \Bbb{R}[x]$. Let $x$ be mapped to $x$ and let $y$ be mapped to $x^2$. We can see why the ideal $(y-x^2)$ goes to $0$. Hence, this map is well-defined. Now we need to construct a morphism between the varieties $V((0))$ and $V(y-x^2)$. We may $t\to (t,t^2)$. In general, if we have a map $\Bbb{C}[x_1,x_2,\dots,x_m]/I\to \Bbb{C}[x_1,x_2,\dots,x_n]/J$, with the corresponding mappings $x_i\to p_i(x_1,x_2,\dots,x_n)$, then the map between the varieties $V(J)\to V(I)$ is defined as follows: $(a_1,a_2,\dots,a_n)\to (p_1(a_1,a_2,\dots,a_n), p_2(a_1,a_2,\dots,a_n),\dots, p_m(a_1,a_2,\dots,a_n))$. How do we know that the image of the point belongs to $V(I)$? Let $f$ be a polynomial in $I$. This is not a difficult argument, and follows from the fact that every polynomial in $I$ maps to a polynomial in $J$ (as the mapping between the coordinate rings is well-defined). We shall try and replicate that argument here.

Let $f\in I$. Then $f(x_1,x_2,\dots,x_m)$ is mapped to $f(p_1(x_1,x_2,\dots,x_n), p_2(x_1,x_2,\dots,x_n),\dots,p_m(x_1,x_2,\dots,x_n))$, which is a polynomial in $J$. This polynomial satisfies the point $(a_1,a_2,\dots,a_n)$. Hence, $f(p_1(a_1,a_2,\dots,a_n), p_2(a_1,a_2,\dots,a_n),\dots,p_m(a_1,a_2,\dots,a_n))=0$. This proves that the map of $(a_1,a_2,\dots,a_n)$ is again a point in $V(I)$, and we have defined a map between the varieties $V(J)$ and $V(I)$.

Where does the isomorphism figure in this picture? One step at a time, young Padawan.

3. Explicit example of a differential form: $(x+y+z)dx+(x^2+y^2+z^2)dy+(x^3+y^3+z^3)dz$. A differential form is just a bunch of functions multiplied with $dx,xy,dz$.

4. Explicit example of the snake lemma in action: I am going to try and talk a bit about Jack Schmidt’s answer [here](http://math.stackexchange.com/questions/182562/intuition-behind-snake-lemma). It promises to be very illustrative.