Irreducibility of Polynomials over $Z/nZ$

Polynomial Factorisation is often a troublesome thing.

Recently, I encountered the topic of irreducibility during my study of ring theory.

Often books treat irreducibility in special rings namely Integral Domains wherein a prime element is an irreducible element and that of Unique Factorisation Domains where the converse also holds true. The definition provided therein is that those elements which cannot be factored as a product of two non-units are called irreducible.

So what happens when we look at rings which are not Integral Domains?

One helpful definition that I encountered is the following :

Let R be a Ring, a polynomial $p(x) \in R[x]$ is said to be irreducible if the ideal generated by $p(x)$ is maximal.

I might not be able to describe the units, but I can always check whether or not the Quotient Ring $R[x]/(p(x))$ is a field.

For now, this seems to be an easier task.

So here's a nice question I encountered during one of my tests:

Q. Is $f(x)=x$ irreducible over $Z/12Z$?

Sol. Observe that $R=Z/12Z$ is not an integral domain, as $4*3 \equiv 0$ mod $12$. So, let's check whether the ideal generated by $x$ is maximal.The elements in $R[x]/(f(x))$ look like $c+A$ where $A=(x)$ is the ideal generated by $f(x)$. Now, since $c\in R$, $c^{-1}$ might not exist. Hence $R[x]/(p(x))$ is not a field.

Thus x is reducible over R.

At first sight, it seems that $x$ is irreducible since we are used to the definitions used in Integral Domains, but on carefully verifying, we find that it is not so.


That's all for this post,
Ciao!

The Devils Staircase

I have been fortunate to encounter the Devil's staircase today. It has shown me that one can perform the devilish act of continuously increasing yet staying constant almost everywhere.

The Devil's staircase also called Cantor's function, is a continuous function $c:[0,1]\rightarrow[0,1]$, with slope zero everywhere except countably many points at which it is one.

Before talking about its construction, let us look at what the construction of the Cantor set as the ideas involved are similar

The Cantor Set

Consider the set $[0,1]$, Now perform :

Step 1: Partition it into 3 parts.

Step 2: Remove the middle third.

Step 3:Repeat Step 1 and 2 on the remaining strips infinitely many points.

We now obtain something like this after six iterations of the above process :

Fig.1

 This set is an example of what is called a perfect set. We will talk of perfect sets in some other article, but for now, observe that this set consists of countably many gaps.

The Staircase

Now that we have seen the cantor set let us finally see what the devil has made for us. 

Consider the straight line $f:[0,1]\rightarrow[0,1]$ such that $f (x)=x$. 

Step 1: Partition it into 3 parts.

Step 2: Take the middle third and make it a horizontal line.

Step 3: Repeat Step 1 and 2 on the remaining part of $f$ infinitely many points.

Call the resulting function $c$. Now $c:[0,1]\rightarrow[0,1]$ is continuous, has slope zero except at countably many points elsewhere one.

Observe the similarity between the cantor set and cantor function, on how at countably many points the $[0,1]$ has gaps and $[0,1]$ has points with slope one and uncountably many with slope 0.

Isn't it amazing how a function could be constant almost everywhere yet have slope one at countably many points?

       fig 2.

The devil's staircase

That's all for now,

Ciao!

Image sources: Wikipedia