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## Month: October, 2013

### Integral domains and characteristics

Today we shall talk about the characteristic of an integral domain, concentrating mainly on misconceptions and important points.

An integral domain is a commutative ring with the property that if $a\neq 0$ and $b\neq 0$, then $ab\neq 0$. Hence, if $ab=0$, then $a=0$ or $b=0$ (or both).

The characteristic of an integral domain is the lowest positive integer $c$ such that $\underbrace{1+1+\dots +1}_{\text{ c times}}=0$.

Let $a\in R$. Then $\underbrace{a+a+\dots +a}_{\text{ c times}}=a\underbrace{(1+1+\dots +1)}_{\text{ c times}}=0$. This is because $a.0=0$.

If $\underbrace{a+a+\dots +a}_{\text{ d, then we have $a\underbrace{(1+1+\dots +1)}_{\text{ d times}}=0$. This is obvious for $a=0$. If $a\neq 0$, then this implies $\underbrace{1+1+\dots +1}_{\text{ d times}}=0$, which contradicts the fact that $c$ is the lowest positive integer such that $1$ added $c$ times to itself is equal to $0$. Hence, if $c$ is the characteristic of the integral domain $D$, then it is the lowest positive integer such that any non-zero member of $D$, added $c$ times to itself, gives $0$. No member of $D$ can be added a lower number of times to itself to give $0$.

Sometimes $\underbrace{a+a+\dots +a}_{\text{ c times}}$ is written as $ca$. One should remember that this has nothing to do the multiplication operator in the ring. In other words, this does not imply that $\underbrace{a+a+\dots +a}_{\text{ c times}}=c.a$, where $c$ is a member of the domain. In fact, $c$ does NOT have to be a member of the domain. It is just an arbitrary positive integer.

Now on to an important point: something that is not emphasized, but should be. Any expression of the form $\underbrace{\underbrace{a+a+\dots +a}_{\text{m times}}+\underbrace{a+a+\dots +a}_{\text{m times}}+\dots +\underbrace{a+a+\dots +a}_{\text{m times}}}_{\text{n times}}=\underbrace{(a+a+\dots +a)}_{\text{m times}}(\underbrace{1+1+\dots +1}_{\text{n times}})$.

Now use this knowledge to prove that the characteristic of an integral domain, if finite, has to be $0$ or prime.

### Ordinals- just what exactly are they?!

If ordinals have not confused you, you haven’t really made a serious attempt to understand them.

Let me illustrate this. If I have 5 fruits (all different) and 5 plates (all different), then I can bijectively map the fruits to plates. However, I arrange the fruits or plates, I can still bijectively map them.

Let’s suppose I have a set finite set $A$, and don’t know its cardinality. But I know hat it bijectively maps to $B$. This directly implies that however I arrange $A$ or $B$, they will still bijectivey map to each other.

This intuition fails for infinte sets. In fact weird things start happening for infinite sets. Natural numbers can be bijectively mapped to rational numbers. What?!! Isn’t the set of rational numbers a superset of natural numbers?! Yes. Then how can there be a bijection between them? Bijection implies both sets contain the same number of elements.

No. That the cardinality should be the same is not part of the definition of bijection. Bijection is defined as an injective and surjective mapping between two sets. It is just that same cardinality is implied through bijection in the case of finite sets. For rational numbers, by cantor’s diagonalization argument, for every number, we cam find a unique pre-image amongst the natural numbers. Hence, we have a bijection.

Coming back to ordinals, let $\omega$ denote the ordered set of natural numbers. Does $\Bbb{N}$ bijectively map to $\Bbb{N}\cup \{\pi\}$? Not if you use the mapping $f(n)=n$. However, if you map $f(1)=\pi$ and $f(n)=n-1$, then you’re done. Note the fact that if two infinite sets are bijective, that does not imply that every one-to-one mapping will be surjective. It just means that there exists *one* such mapping. This is in direct contrast with the case of finite sets, in which every injective mapping between two bijective sets is surjective.

What if you map $\omega$ to $\omega+1$? Note that as $\omega$ is ordered, this implies there is only *one* mapping we’re allowed to have: $f(n)=n$. We can clearly see $\Bbb{N}$ can’t be bijectively mapped to $\omega+1$.

Where most mathematical texts fail is actually explaining these finer points to students. Most of them just regurgitate the material present in “classics” of that subject. The most important thing to note here is that if we have two infinite sets which are not ordered, then there *might* be some bijective mapping between them, and finding one can be tricky sometimes. For example, Cantor’s diagonal mapping is brilliant, and non-trivial. Hence for non-ordinal infinite sets, we can’t be sure if they’re bijective with $\Bbb{N}$. However, in the case of ordinals, as there is only one mapping, determining whether the set is bijective with respect to $\Bbb{N}$ is trivial.