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Understanding Irreducibles

Bingk

New member
Jan 26, 2012
16
Hi,
I'm trying to get a better understanding of irreducible elements.

Obviously, the product of two irreducibles is reducible, by definition.
I think the product of two reducibles is also reducible (I'm under the assumption that a reducible is also a non-zero, non-unit element, and hence the product of two reducibles will be a product of two non-units).
Also, the product of an irreducible and reducible should also be reducible (again, a product of two non-units).
This leads me to the idea that reducibility is basically a way of categorizing non-zero, non-units, where irreducible elements are like prime elements of non-units (i.e., a non-zero, non-unit element can be factored into a product of irreducibles). Is it okay to think of it this way?

I'm wondering if we can say anything about the product of the following:
a) an irreducible and a unit;
b) a reducible and a unit;
c) an irreducible and a non-unit;
d) a reducible and a non-unit.

I think for b) it will still be reducible, since it is the product of two non-units, when it is multiplied to/by a unit, you will still get a product of two non-units (i.e. the unit is multiplied to one of the non-units, still resulting in a non-unit).
For a), I think it will still be irreducible. If it is written as the product of two units, then multiplying another unit to either of the units will still result in a product of two units (and thus cannot be the product of two non units). If it is written as a product of a non-unit and unit, then if the unit is multiplied to either the unit or non-unit, we still get a product of a non-unit and unit.
For d), I think it will still be reducible, since a product of two non-units is a non-unit, so when you have a product of three non-units, you can reduce it to a product of two non-units.
for c) I think it depends. If it cannot be written as the product of a unit and non-unit (i.e. can only be written as a product of units), the it will still be irreducible, as one of the units will remain when the other is multiplied to the non-unit. But, if it can be written as the product of a unit and non-unit, it will become reducible since the product of a unit and non-unit will be a non-unit, so when multiplied to the other non-unit, it will be a product of two non-units.

Any comments/corrections to my thinking?
Thanks in advance! :D
 

Swlabr

New member
Feb 21, 2012
27
Hi,
I'm trying to get a better understanding of irreducible elements.

Obviously, the product of two irreducibles is reducible, by definition.
I think the product of two reducibles is also reducible (I'm under the assumption that a reducible is also a non-zero, non-unit element, and hence the product of two reducibles will be a product of two non-units).
Also, the product of an irreducible and reducible should also be reducible (again, a product of two non-units).
This leads me to the idea that reducibility is basically a way of categorizing non-zero, non-units, where irreducible elements are like prime elements of non-units (i.e., a non-zero, non-unit element can be factored into a product of irreducibles). Is it okay to think of it this way?

I'm wondering if we can say anything about the product of the following:
a) an irreducible and a unit;
b) a reducible and a unit;
c) an irreducible and a non-unit;
d) a reducible and a non-unit.

I think for b) it will still be reducible, since it is the product of two non-units, when it is multiplied to/by a unit, you will still get a product of two non-units (i.e. the unit is multiplied to one of the non-units, still resulting in a non-unit).
For a), I think it will still be irreducible. If it is written as the product of two units, then multiplying another unit to either of the units will still result in a product of two units (and thus cannot be the product of two non units). If it is written as a product of a non-unit and unit, then if the unit is multiplied to either the unit or non-unit, we still get a product of a non-unit and unit.
For d), I think it will still be reducible, since a product of two non-units is a non-unit, so when you have a product of three non-units, you can reduce it to a product of two non-units.
for c) I think it depends. If it cannot be written as the product of a unit and non-unit (i.e. can only be written as a product of units), the it will still be irreducible, as one of the units will remain when the other is multiplied to the non-unit. But, if it can be written as the product of a unit and non-unit, it will become reducible since the product of a unit and non-unit will be a non-unit, so when multiplied to the other non-unit, it will be a product of two non-units.

Any comments/corrections to my thinking?
Thanks in advance! :D
So, an irreducible element of an integral domain is a non-zero element which is not the product of two non-units.

You are right in thinking that the product of two irreducibles is reducible - this is basically the definition of reducible!

Unfortunately, it is not okay to think of irreducibles as prime elements, although in some ways they are. This is because integral domains have prime elements! Prime elements are to do with division: $a$ is prime if $a|bc$ then $a|b$ or $a|c$.

The product of an irreducible and a unit is also irreducible, as if $ab=uc$ where $u$ is a uni $c$ is irreducible and $a$ and $b$ are non-units, then $u^{-1}ab=c$ where $u^{-1}a$ and $b$ are both non-units, a contradiction.

A reducible and a unit will still be reducible: $abu=a(bu)$ where $a$ and $b$ are both non-units. It sufficies to prove that $bu$ is also a non-unit. If it was, then the product of two units is a unit so $b=buu^{-1}$ is a unit, a contradiction.

An irreducible with a non-unit will be reducible, by definition, as irreducibles are non-units. The same argument works for (d).

The problem with your argument for (c) is, I think, you are forgetting that irreducibles are explicitly not allowed to be units.