# Understanding Irreducibles

#### Bingk

##### New member
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.

#### Swlabr

##### New member
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.

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.