$x^2, x^3$ being irreducible in $F[x^2, x^3]$

Question

(For $F$ field) $F[x^2,x^3]$$\;$clearly $x^2$ is irreducible as$\;$ $x^2=x\cdot x$$\;$ but $x\not\in F[x^2,x^3]$.

Can the same be applied to $x^3?$$\;$ $x^3=x^2\cdot x$$\;$ (Neither of which are units.)

If I am totally incorrect, how do I show that $x^2,x^3$ are irreducible in $F[x^2, x^3]$,

Thanks for the help!

Answer 1

Definition of irreducible element in any ring $R$ is as follows: $p$ is irreducible in $R$ if $$p=ab \implies a\in R^\times\ \text{or}\ b\in R^\times;\quad a,b\in R$$

You are free to think of irreducible elements as those that have exactly two divisors (up to multiplication by unit) - $1$ and themselves. That means that if you want to prove that $x^2$ is irreducible in $F[x^2,x^3]$, you need to show that it is impossible to write $x^2 = p(x)q(x)$ in $F[x^2,x^3]$, unless $p(x)$ or $q(x)$ is unit.

If you write $x^2 = x\cdot x$, this is not factorization in $F[x^2,x^3]$, so this tells you nothing of irreducibility in $F[x^2,x^3]$ (at least not directly). In contrast, it is factorization in $F[x]$ and is unique factorization into irreducibles in UFD $F[x]$.

You can proceed by assuming that $x^2 = p(x)q(x)$, where $p(x),q(x)\in F[x^2,x^3]$ and assume that neither is unit. But that means that $\deg p, \deg q \geq 1$, which implies $\deg p = \deg q = 1$, since $\deg p + \deg q = 2$. But, this is contradiction, since there are no polynomials of degree $1$ in $F[x^2,x^3]$. Similarly for $x^3$.

Answer 2

You should rely on unique prime factorization in $F[x]$.

Your factorization $x^3 = x^2 \cdot x$ is of course irrelevant because $x \notin F[x^2,x^3]$. It's like saying $3 = 9\cdot\frac13$ and calling that a factorization in $\mathbb Z \subset \mathbb Q$.

Answer 3

Your best friend is the degree. $\deg(fg)=\deg(f)+\deg(g)$.