Is the composition of irreducible polynomials again irreducible? I've been pondering this since yesterday. I

Is it true that given two irreducible polynomials $f(x)$ and $ g(x)$ will $f(g(x))$ or $g(f(x))$ be irreducible?

 A: I am extending @Pipicito's answer for the case when both polynomials have degree greater than 1. The answer is also no. For example, consider $f(x)=x^2-\frac{4}{3}$; it is irreducible over $\mathbb{Q}$. However, $f(f(x))={(x^2-\frac{4}{3})}^2-\frac{4}{3}=(x^2-2x+\frac{2}{3})(x^2+2x+\frac{2}{3})$ and it is reducible!
Moreover, there is Cappelli's lemma about this question. It says the following:
Let $K$ be a field, $f$ and $g$ be irreducible polynomials over $K$.  Then, $f(g(x))$ is irreducible over $K$ iff $f$ is irreducible over $K$ and $g-\alpha$ is irreducible over $K(\alpha)$ for every root $\alpha$ of $f$.
In the example, the roots of $x^2-\frac{3}{4}$ are $\pm \frac{\sqrt{3}}{2}$.
So, to show $f(f(x))$ is reducible, just show that one of $x^2-\frac{3}{4}\pm \frac{\sqrt{3}}{2}$ has roots in $\mathbb{Q}(\sqrt{3})$
A: I don't know a good answer to this question. But here are a couple of simple ideas for easy cases.
1) If $\mathbb{K}$ is a field then for any two irreducible polynomials $f$ and $g$ in $\mathbb{K}[x]$ such that $g$ has degree one we have that $f \circ g$ is irreducible. To see this, write $g=ax+b$ with $a\neq 0$. Then if $f \circ g$ were reducible you would have $p$ and $q$ in $\mathbb{K}[x]$, both of them with degree at least one, with $f\circ g= p\cdot q$. But as $\mathbb{K}$ is a field you can find the "inverse" of $g$, call it $t=a^{-1}x-ba^{-1}$. So now $ f = f \circ x = f\circ( g \circ t) = (f\circ g) \circ t = (p\cdot q) \circ t = (p\circ t)\cdot (q \circ t)$. But this contradicts the irreducibility of $f$ since $p\circ t$ and $q \circ t$ have degree at least one.
2) If in 1) we replace $\mathbb{K}$ with an arbitraty ring then it's no longer true. For example in $\mathbb{Z}[x]$ you can take $f=5x+7$ and $g=2x+3$. Both of them are reducible but $f \circ g = 2\cdot (5x+11)$. Note that in $\mathbb{Z}[x]$ the polynomial $2$ is not a unit.
3) If in 1) we take $f$ to be of degree one then it's no longer true. Even more, if $g$ is any polynomial of degree two or more we can always find a polynomial $f$ of degree one such that $f\circ g$ is reducible. Since $g$ has degree two or more we can write $g= x\cdot h + a$, $a$ is the "constant term" of $g$. Now took $f=x-a$. It turns out that $f\circ g= x \cdot h$. This last equality gives a true factorization since the degree of $f\circ g$ is the same as the degree of $g$ because $f$ has degree one and $\mathbb{K}$ is, in particular, an integral domain. Note that the reasoning works fine in any integral domain.
A: This need not be true. Note that in $\mathbb{R}[x]$ any irreducible polynomial has degree either $1$ or $2$. So, you can take two irreducible polynomials of degree 2, and compose them to get a reducible polynomial.
