Units, Primes and Irreducibles

Question

How do you find the units, irreducible elements and prime elements for $\mathbb{C}[𝑥]$, $\mathbb{R}[𝑥]$, $\mathbb{Q}[𝑥]$?

Thank you.

Answer 1

In general, if $A$ is a commutative ring with $1$, then $p=p_0+xp_1+\cdots+x^mp_m\in A[x]$ is a unit if and only if $p_0\in A^*$ and $p_j$ is nilpotent for all $j\ge 1$. Specializing this lemma is trivial.

The irreducible elements of $\Bbb C[x]$ are exactly the non-zero primes of $\Bbb C[x]$. By a very deep result due to Gauss, they are exactly the polynomials of degree $1$.

The irreducible elements of $\Bbb R[x]$ are exactly the non-zero primes of $\Bbb R[x]$. By the aforementioned deep result, they are exactly the polynomials which either have degree $1$ or are in the form $ax^2+bx+c$ for some $a,b,c\in\Bbb R$ such that $a\ne 0$ and $b^2-4ac<0$.

The irreducible elements of $\Bbb Q[x]$ are exactly the non-zero primes of $\Bbb Q[x]$ and they are exactly ¯\_(ツ)_/¯.

Answer 2

These examples are all of the form $K[x]$, where $K$ is a field.

The units of $K[x]$ are exactly $K^*$, the nonzero elements of $K$.

Since $K[x]$ is a PID, irreducible and prime are the same concept.

The irreducible elements of $K[x]$ depend deeply on the arithmetic properties of $K$.

For $K=\mathbb C$ and $K=\mathbb R$, the irreducible elements of $K[x]$ follow from the fundamental theorem of algebra:

• The irreducible elements of $\mathbb C[x]$ are the polynomials of degree $1$.

• The irreducible elements of $\mathbb R[x]$ are the polynomials of degree $1$ or the polynomials of degree $2$ with negative discriminant (that is, with nonreal roots).

For $K=\mathbb Q$, there is no simple answer:

• The irreducible elements of $\mathbb Q[x]$ correspond to the minimal polynomials of algebraic numbers. There is no simple description of those.