Let $P \subset \mathbb R[x]$ be defined as follows:
$$P = \{ax^3+bx+c \mid \ a,b,c \in \mathbb R\}$$
then let the operation $*$ be so defined on $P$:
$$(ax^3+bx+c)*(ux^3+tx+z) = aux^3+btx+cz$$
Then find out
- whether $(P,*)$ is a monoid and if that's the case determine its invertible elements;
- for which pairs $(f,g) \in P \times P$ the polynomial $f * g$ is not reducible in $\mathbb (R, +, \cdot)$.
The structure $(P,*)$ is a monoid if the operation $*$ is associative and if the struction has an identity element. So, let $f,g,h \in P$, $*$ is associative if
$$(f*g)*h = f*(g*h)$$
that is
\begin{equation} \left((ax^3+bx+c)*(ux^3+tx+z)\right)*(wx^3+sx+k) =\\ (aux^3+btx+cz) * (wx^3+sx+k) = \\ (auwx^3+btsx+czk) = \\ (ax^3+bx+c) * (uwx^3+tsx+zk) = \\ (ax^3+bx+c) * \left( (ux^3+tx+z) * (wx^3+sx+k)\right) \end{equation}
So $*$ is associative and since the identity element of the structure is the polynomial $x^3+x+1$, then $(P,*)$ is a monoid. The invertible elements of this structure are all the polynomials $ax^3+bx+c$ such that
$$(ax^3+bx+c) * (\alpha x^3 + \beta x + \gamma) = x^3+x+1$$
and that happens as
\begin{cases} a \cdot \alpha = 1 \Leftrightarrow \alpha = a^{-1} \\ b \cdot \beta = 1 \Leftrightarrow \beta = b^{-1} \\ c \cdot \gamma = 1 \Leftrightarrow \gamma = c^{-1} \end{cases}
and clearly there is no inverse for polynomials $f \in P$ such that $a = 0 \vee b = 0 \vee c = 0$.
I am struggling with the second request, I don't know how to find a polynomial $f*g$ so that it is irreducible in $(\mathbb R, +, \cdot)$. Any help, suggestion, push in the right direction is very welcome.
