Units, Primes and Irreducibles How do you find the units, irreducible elements and prime elements for $\mathbb{C}[]$, $\mathbb{R}[]$,  $\mathbb{Q}[]$?
Thank you.
 A: 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)_/¯.
A: 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.

