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.

  • 2
    $\begingroup$ It is not true that if $p$ is reducible, then it is a unit. What makes you think this? $\endgroup$ – Servaes Apr 2 at 16:31
  • $\begingroup$ 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? $\endgroup$ – Kbiir Apr 2 at 16:35
  • $\begingroup$ 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. $\endgroup$ – hardmath Apr 2 at 16:50
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    $\begingroup$ @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). $\endgroup$ – Jose Brox Apr 2 at 18:21
  • $\begingroup$ @JoseBrox you are absolutely right. I should be more carful. $\endgroup$ – 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.


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