Vrftjkry

The definition claims that a polynomial in a field of positive degree is a reducible polynomial when it can be written as the product of $2$ polynomials in the field with positive degrees. Other wise it is irreducible.

So if a polynomial $f(x)$ can be written as the product of say $41(x^2 + x)$, is that considered not reducible because $41$ is really $41x^0$, and $0$ isn't technically positive, but by the definition of a polynomial in a field it is a polynomial if $a_n$ isn't $0$ for the highest degree $n$ where $n geq 0$.

So would the example of the polynomial example I gave be reducible or irreducible?

This particular example would be reducible, but not because of the reason you give -- it would be because you could write it as $41x(x+1)$. Note that $0$ is not positive, so the factorization you give is not a product of two polynomials with positive degree ($41$ is a polynomial of degree $0$).

Definition : Given an integral domain $R$, a non-zero non-unit element $r in R$ is said to be irreducible if it cannot be written as a product of non-units i.e. whenever it is written as a product of two elements, at least one of them is a unit in $R$.

For polynomials, this becomes :

Definition : Given an integral domain $R$, the ring $R[X]$ also is an integral domain, and $f$ is an irreducible polynomial over $R$ if it is an irreducible element of $R[X]$.

So, it is that simple. Let us take some examples to clarify.

• The polynomial $f(x) = x$ is irreducible over any ring, since if $a(x)b(x) =x$, then WLOG $a$ must be a constant polynomial with the constant being a unit(use the rules for multiplication of polynomials), so $a$ is a unit in $R[X]$, hence $x$ is irreducible.




• The polynomial $f(x) = 2x+2$ is irreducible over $mathbb R[X]$. This is because if $a(x)b(x)$ divides $2(x+1)$ then at least one of $a(x)$ or $b(x)$ is a constant polynomial, but every constant polynomial is a unit in $mathbb R$. However, this polynomial is reducible over $mathbb Z[X]$, since here, $2(x+1)$ counts as a non-unit factorization, because $2$ is not a unit.

Therefore, reducibility depends on "over which ring/field"? For example, $41 = 41x^0$ is a constant polynomial, but it isn't a unit in $mathbb Z[X]$, while it is one in $mathbb R[X]$. So, a polynomial like $41(x+1)$ is irreducible over the latter but not over the former.

While working over a field, it turns out that the set of units of $F[X]$ is equal to the non-zero constant polynomials. Therefore, any polynomial is irreducible in $F[X]$ if and only if it can be written as the product of two non-constant polynomials. Things will change for a ring which is not a field, since some non-constant polynomials may possibly not be units.

Also, note that $41(x^2+x)$ is reducible in every field where it is non-zero, since it can be written as $(41 x) times (x+1)$ which is the product of two non-constant polynomials, which are always non-units by the fact that the degree is multiplicative.

The reason why "where it is non-zero" is mentioned, is because in characteristic $41$(or, over fields of characteristic $41$ such as $mathbb Z over 41mathbb Z$), this polynomial actually reduces to $0$, for which the notion is irreducibility is not defined above. For fields of any other characteristic, the polynomial is non-zero(and non-unit) so we can discuss irreducibility. Much thanks to the comment below for pointing this out. Also, something similar applies to other polynomials discussed previously such as $2x+2$ (in characteristic $2$, this polynomial is identically zero) and $41(x+1)$. However, $x$ remains non-zero over every field, so the argument remains.

A polynomial $f(x)$ is reducible if $f(x)=g(x)h(x)$ with $deg(g(x)) ge 1$, and $deg(h(x)) ge 1$