# polynomial units in field

I am new on this topic and would like to have some clarifications.

We were given a definition and theorem as follows:

Definition : A polynomial $$p$$ in $$F[x]$$ is irreducible if $$p$$ is not a unit of $$F[x]$$, and if $$p = f\cdot g$$ then either $$f$$ or $$g$$ must be a unit.

By my own interpretation if it is reducible then $$p$$ must be a unit of $$F[x]$$.

Theorem: If $$F$$ is a field then the only units of $$F[x]$$, that is polynomials $$p$$ such that exists $$q$$, $$p\cdot q = 1$$ and in ($$F[x]$$), are the units of $$F$$. Thus it can only be constant polynomial of degree $$=0$$.

Then we have this fact $$\dots$$ $$(x^2)+1$$ is irreducible in $$ℤ/3ℤ$$ (but not in $$ℤ/5ℤ$$!)

Now my professor said it is irreducible because it cannot be factored in $$ℤ/3ℤ$$ but can in $$ℤ/5ℤ$$ which is $$(x+3)\cdot (x+2)$$. I am confused because the theorem stated that units of $$F$$ are only constant numbers. However we have $$(x^2)+1$$ which is reducible in $$ℤ/5ℤ$$ and thus is a unit in $$ℤ/5ℤ$$???

I am so confused, I don't know where or what I am doing wrong.

• It is not true that if $p$ is reducible, then it is a unit. What makes you think this? – Servaes Apr 2 at 16:31
• I was going with the definition... if p is irreducible then it is not a unit of F[x]. so I though the inverse is also true, if p is reducible then it is a not in F[x]. If I may ask why is it not true? – Kbiir Apr 2 at 16:35
• One defines irreducible by requiring more than just not being a unit. It is stated in the Question, if polynomial $p = f\cot g$ then either $f$ or $g$ is a unit. – hardmath Apr 2 at 16:50
• @Kbiir That is a usual logical flaw: if you know that fact $A$ implies fact $B$, then you CANNOT infer that the negation of $A$ implies the negation of $B$. For example: from "if what I'm seeing is a tomato then it has red color" you cannot infer "if what I'm seeing is not a tomato then it does not have red color"! (It could be an strawberry, or a red car, or whatever). From $A\Rightarrow B$ you cannot infer $\neg A\Rightarrow \neg B$... but you can infer the contrapositive, $\neg B\Rightarrow \neg A$ (if it is not red, it is not a tomato). – Jose Brox Apr 2 at 18:21
• @JoseBrox you are absolutely right. I should be more carful. – Kbiir Apr 3 at 2:47

It is not true that if $$p$$ is reducible then it is a unit. Recall that an element $$p\in F[X]$$ is a unit if there exists $$q\in F[X]$$ such that $$pq=1$$. If $$F$$ is a field, then comparing degrees shows that all units are nonzero constants. Conversely all nonzero constants are clearly units.

Now consider for example the polynomial $$p=X^2\in F[X]$$, which is not a unit. Then for the polynomials $$f=g=X\in F[X]$$ we have $$p=fg$$, and neither $$f$$ nor $$g$$ is a unit. This means $$p$$ is reducible, but not a unit.

The same reasoning shows that more generally, a product of two nonconstant polynomials in $$F[X]$$ is reducible but not a unit.

Here $$F$$ is assumed to be a field for the purpose of your Theorem. In that case the units of $$F[X]$$ are the nonzero constants (i.e. units of $$F$$), where $$X$$ is assumed to be an indeterminate.

More generally if $$F$$ is an integral domain (no zero divisors), then the units of $$F[X]$$ are again the units of $$F$$ (considered as degree zero polynomials).

The point of defining "irreducibles" to exclude units is to give a definition that is widely applicable. We could have said, apropos of a polynomial ring, that the degree should be nonzero, but the way stated applies to "irreducibles" in any integral domain.

For univariate polynomials with coefficients over a field, an irreducible polynomial is a nonzero polynomial that cannot be factored except as a unit times another polynomial. This also generalizes to polynomials in more than one variable.