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Month: March, 2017

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.

Irritating set of examples- I

I am trying to collect explicit examples for concepts and calculations. My hope is that this website becomes a useful repository of examples for anyone looking for them on the internet.

First some words of wisdom from the master himself, Professor Ravi Vakil: “Finally, if you attempt to read this without workign through a significant number of exercises, I will come to your house and pummel you with the EGA until you beg for mercy. As Mark Kisin has said, “You can wave your hands all you want, but it still won’t make you fly.”

We first start with some category theory examples:

1. Can we have two products of the same two objects, say $A$ and $B$, in the same category? This question is much more general than I am making it out to be. Can we have two distinct universal objects of the same kind in a category (although they may be isomorphic, and even through unique isomorphism)? The only example of the product of objects being isomorphic but not the same is the following: $A\times B$ and $B\times A$. These aren’t the same objects, but they’re isomorphic through unique isomorphism.

2. Groupoid- In the world of categories, a groupoid is a category in which all morphisms between objects are isomorphisms. An example of a groupoid, which is not a group, is the category $\mathfrak{Set}$ with the following restriction: $\text{Hom(A,B)}$ now only consists of isomorphisms, and not just any morphisms. This example, although true, is not very illustrative. This [link](http://mathoverflow.net/questions/1114/whats-a-groupoid-whats-a-good-example-of-a-groupoid) provides a much better demonstration of what is going on. Wikipedia says that the way in which a groupoid is different from a group is that the product of some pairs of elements may not be defined. The Overflow link suggests the same thing. You can’t take any any pair of moves that one may make on the current state of the jigsaw puzzle, and just compose them. The most important thing to note here is that the elements of the group do not correspond to objects of the categories. They correspond to morphisms between those objects. This is the most diabolical shift of perspective that one encounters while dealing with categories. Suddenly, morphisms encode much more information than you expect them to.

3. Algebraic Topology example: Consider a category in which points are objects of the category, and the paths between points, upto homotopy, are morphisms. This is a groupoid, as paths between points are invertible. The return path should not wrap around a wayward hole, obviously. One may consider the path as the same, just travelling in the opposite direction. The automorphism group of a point would be the fundamental group of paths centred at that point.

Another category that stems from Algebraic Topology is one in which all objects are topological spaces, and the morphisms between maps are the continuous maps between those spaces. Predictably, the isomorphisms are the homeomorphisms.

4. Subcategory: An example would be one in which objects are sets with cardinaly $1$, and morphisms would be the same as those defined in the parent category- $\mathfrak{Set}$.

5. Covariant functor: Consider the forgetful functor from $\mathfrak{Set}$ to $\mathfrak{Vec}_k$. The co-domain is bigger than the domain. One could think of this functor as an embedding.

A topological example is the following: one which sends the topological space $X$, with the choice of a point $x_0$, to the object $\pi(X,x_0)$. How does this functor map morphisms? It just maps paths in $X$ to their image under the same continous map. How do we know that the image is a path? This is easy to see. We can prove that we ultimately have just a continuous map from $[0,1]$ to that image, and we will be done. Do we have to choose a point in each topological space? Yes. What if we have the following two tuple $(X,x_0), (Y,y_o)$, such that $x_0$ is not mapped to $y_0$? Then there is no morhism between these two objects. In other words, the set of morphisms $\text{Hom}((X,x_0), (Y,y_o))$ consists of only those morphisms which map $x_0$ to $y_0$. An illustrative example is the following: $f_1: [0,1]\to (\cos (2\pi t),\sin (2\pi t))$ and $f_2: [0,1]\to (\cos (4\pi t),\sin (4\pi t))$. These are two different continuous maps between the same two topological spaces. They both map $0$ to the point $(1,0)$ in $S^1$, but they map a path starting and ending

Side note: Example of two homotopic paths being mapped to homotopic paths under a continuous map. Let $f: [0,1]\to S^1$ be the continuous map under consideration. Consider any path $p$ in $[0,1]$ which starts and ends at $0$. We know that this is homotopic to the constant path at $0$ (one may visualize the homotopy as shrinking this path successively toward $0$). Then the image of this homotopy is mapped to a path in $S^1$ that shrinks toward the constant map at $(1,0)$.

6. Contravariant functors: Mapping a vector space to its dual. This example is pretty self-explanatory.

7. Natural Transformation: A natural transformation is a morphism between functors. Abelianization is a common example of a natural transformation. The two functors, both of which are covariant, are $id$ and $id^{ab}$. The first one maps a group to itself, and the second one maps a group to its commutator. The resultant commutative diagram is easy to see too. The data of the natural transformation is just $m:G\to G^{ab}$ and $m:G'\to G'^{ab}$.

The double dual of a vector space is another example of a natural transformation. The dual would have worked too, except for the fact that the dual functor is contravariant. Note: one of the functors, in both these natural transformations, is the identity functor.

8. Equivalence of categories- This is exactly what you think it is. Two categories that are not equivalent are $\mathfrak{Grp}$ and $\mathfrak{Grp^{ab}}$. Too much information is lost while abelianizing the group, which cannot be regained easily.

9. Initial object- The empty set is the initial object in the category $\mathfrak{Set}$. Why not a singleton? Because the map from the initial object to any object also has to be unique. Moreover, a singleton will not map to an object- namely the empty set. And an initial object should map to all objects.

10. Final object- A singleton will be a good final object in the category $\mathfrak{Set}$.

11. Zero object- The identity element in the category $\mathfrak{Grp}$ would be such an object.

12. Localization through universal property: Consider $\Bbb{Z}$, with the multiplicative subset $\Bbb{Z}-\langle 7\rangle$. The embedding $\iota: \Bbb{Z}\to \Bbb{Q}$ ensures that every integer goes to an invertible element. Trivially, so does every element of $\Bbb{Z}-\langle 7\rangle$. Hence, there exists a unique map from $(\Bbb{Z}-\langle 7\rangle)^{-1}\Bbb{Z}$ to $\Bbb{Q}$. We can clearly see that this is overkill. Many more elements than just those of $\Bbb{Z}-\langle 7\rangle$ are mapped to invertible elements. The point is that there may be a ring $A$ such that only elements of $\Bbb{Z}-\langle 7\rangle$ are mapped to invertible elements in $A$. Hence, in that case too, there will exist a unique map from $(\Bbb{Z}-\langle 7\rangle)^{-1}\Bbb{Z}$ to $A$. Why do we care about there existing a map from some other object to rings which $\Bbb{Z}$ maps to at all? When we have a morphism $\phi:A\to B$, and we can say that there exists a map $A/S\to B$, where $S$ is a set of relations between elements of $A$, then we’re saying something special about the properties of elements in $B$ (at least the properties of elements mapped to by $S$).