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!
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!
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