# Can anyone help to me conceptualize what is meant by irreducible polynomial

Hello I am just in an intro to algebra course and I am having some problems completely understanding what is meant by irreducible polynomial.

Partially this is because I actually have two definitions in my notes, which I am sure are equivalent, but it is just making me confused.

One definition is that first we define the equivalence relation ~ on polynomials in a field by f(x)~g(x) iff $f(x)=ag(x)$ for some non zero $a$ in the field.

then we say f is irreducible if g|f implies that g ~ 1 or g~f.

And second is that we say a non constant polynomial is irreducible in a field $F[X]$ if $f=qs$ then either q or s are units.

I'm just not exactly sure I understand what is going on, like what it is really saying. I don't have any examples either, and am not sure which definition makes more sense etc. Hopefully someone can help with my understanding.

Thank you

• $x^2-1$ can be factored into $(x-1)(x+1)$, so it's not irreducible. $x^2+1$ cannot be factorized (over the reals, at least), so it is irreducible. Basically, irreducible is like "prime" but for polynomials. – Akiva Weinberger Oct 22 '15 at 20:52
• So similar to primes, if we know say that a polynomial f is irreducilble, can we conclude that only plus and minus 1 divide it? or – Quality Oct 22 '15 at 20:55
• Well, $2$ divides $x^2+1$, since $x^2+1=(2)(\frac12x^2+\frac12)$. – Akiva Weinberger Oct 22 '15 at 20:56
• Sorry i meant the only monics – Quality Oct 22 '15 at 20:58
• $\frac12x^2+\frac12$ divides it, and so does $2$ — however, $\frac12x^2+\frac12\sim x^2+1$ and $2\sim1$. As you said in your post, a polynomial $f$ is irreducible iff $g|f$ implies $g\sim f$ or $g\sim1$. (Note that $"\!g\sim1\!"$ is the same as saying "$g$ is a constant.) – Akiva Weinberger Oct 22 '15 at 20:59

Both definitions are equivalent, since in the first definition, $g\sim 1$ simply means $g$ is a non-zero constant, and in a polynomial ring over a field, units are exactly the non-zero constants.
The second definition is slightly more general, in the following sense: if you consider polynomial rings over an integral domain, say $\mathbf Z[x]$, it will be fine, because over an integral domain, being a unit is not the same as being non-zero: in $\mathbf Z$, units are $\pm 1$, hence a polynomial will be irreducible over $\mathbf Z$ if $f=qs$ implies one of $q, s$ is equal to $\pm1$. Thus $7x^2+z$ is not irreducible in $\mathbf Z[x]$ because $5x^2+5=5(x^2+1)$ and neither $5$ nor $x^2+1$ is a unit. However it is irreducible in $\mathbf Q[x]$ because $5$ is unit in $\mathbf Q$.
I would say irreducible polynomials are analogous to what people would tell you prime numbers are. If you ask a person what is a prime number, they invariably will respond by saying $p$ is prime if its only divisors are $1$ and $p$ itself. If they are a math person, they might say $\pm 1$ or $\pm p$. This is really the idea of irreducible and not prime. In $\mathbb{Z}$ it turns out that prime ($p\mid ab \Rightarrow p \mid a$ or $p \mid b$) and irreducible are the same, so it is no big deal, but I find it causes problems in other rings which are not unique factorization domains (UFDs).
This is how I would think about irreducible polynomials: they can only be factored in uninteresting ways: a unit times a unit multiple of the polynomial. Just like $5=(-1)(-5)$ isn't interesting, $p(x)=\frac{1}{2}(2p(x))$ is not interesting, so we would not count this as being a factorization which would prevent $p(x)$ from being irreducible.
Now if you are looking at $K[X]$ where it is polynomials over a field, this is also a UFD, so again prime and irreducible are the same. In general, you can still have irreducible and prime polynomials in the ring $R[X]$, but these may be different. In general, prime implies irreducible but not conversely.