Definition of irreducible polynomial via the units of an integral domain.

Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be irreducible over $D$ if, whenever $f(x)$ is expressed as a product $f(x)= g(x)h(x)$, with $g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$.

This is the definition of the irreducible polynomial in Gallian's book. What I don't understand is how this definition translate to the fact that the polynomial cannot be factored into a product of polynomials of lower degree when $D$ happens to be a field.

There is also this examples that the polynomial $f(x)=2 x^2+ 4$ is irreducible over $\mathbb R$, but reducible over $\mathbb{C}$.

To me it seems that it is also irreducible over $\mathbb{C}$ since in its factorization $2(x^2+2)$, $2$ is unit over $\mathbb{C}$.

• There are two things: If the polynomial satisfies this but can also be factored into two polynomials both of which have smaller degrees, neither of those can be a unit, as those have degree $0$. The other direction is not true. Just because a polynomial cannot be factored into smaller degrees, it need not be irreducible, as for example the polynomial $2x$ over the integers shows. Jun 30 '16 at 9:37
• @TobiasKildetoft: "factored into two polynomials both of which have smaller degrees, neither of those can be a unit, as those have degree 0." why can't either of the factors be unit, (why its degree has to be zero)? can you explain it a bit with any examples?
– MUH
Jun 30 '16 at 9:54
• In your example, remember that the definition says that for all factorizations, one factor must be a unit. Jun 30 '16 at 10:03
• To me it seems that it is also irreducible over ℂ But $x^2+2=(x-i\sqrt{2})(x+i\sqrt{2})$ over ℂ Jun 30 '16 at 10:42

That common definition of an (ir)reducible polynomial works over a field but needs to be modified when you pass to more general coefficient rings, because they may contain constants that are nonzero nonunits. In your example, $\,f = 2x^2+4\,$ is irreducible over $\,\Bbb Q,\,$ but is reducible over $\,\Bbb Z\,$ where $\,f = 2(x^2+2)\,$ and both factors are nonzero nonunits, but both factors are not of lower degree.

Over non-fields, the constants can convey very useful divisibility information so we don't want to ignore them when studying polynomial divisibility. For example, one way to view Gauss's Lemma is that the prime $\,p\in\Bbb Z\,$ extends to a prime in $\,\Bbb Z[x],\,$ i.e. $\,p\mid fg\,\Rightarrow\,p\mid f\,$ or $\,p\mid g.\,$

• but in C this would be: 2(x−i2–√)(x+i2–√), but the 2 would be a unit in C, then why is it still reducible as 2 is a unit?
– user304120
Jun 27 '19 at 18:05
• @user3892683 It's a product of the nonzero nonunits $\,2(x-\sqrt{-2})\,$ and $\,x+\sqrt{-2}\,$ so it is reducible. Jun 27 '19 at 18:13

The basic idea is that for any possible combination of factors if none of them is non zero and non unit then the polynomial is reducible over that integral domain. In ID z(X)only the constants 1 and -1 are units. So 2(x^2)+2 can be factored as (2)((x^2)+1) both non zero and non unit ( although degree of both is not lower). So it is reducible in z(X).

But in case of fields all constants, non zero are units. So to make combination of non zero and non unit factors the degree of both factors must be lower (any possible case) to show the polynomial is reducible. If no such combination exists then polynomial is irreducible. So 2(x^2)+2 if factored as (X+i)(2x-2i) gives terms not belonging to R(X). And if factored as (2)(x^2 +1) then 2 is a unit in R(X). In other words it can't be broken into two factors of lower positive degrees. So it is irreducible in R(X).

In C(X) there exists combination (X+i)(2x-2i) both nonzero, non unit and positive lower degrees. So reducible over c(X).