For $n\ge 1$, let $g(x\_1,x\_2,\ldots,x\_n)$ be an irreducible *homogeneous* polynomial in $n$ variables over a field $k$ and $f(x)$ an irreducible polynomial of $k[x]$. Is $f(g(x\_1,x\_2,\ldots,x\_n))$ necessarily irreducible?


For instance this holds when $n=1$ (since then $g(x\_1)=\lambda x\_1$), or when $f$ has degree 1 (by a simple argument).



Irreducibility of a polynomial
I believe that the answer is yes. Put $c:=g(x\_1,\ldots, x\_n)$, which is irreducible in the UFD $R:=k[x\_1,\ldots, x\_n]$. Assume $f$ is irreducible, but also assume by way of contradiction that $f(c)=\alpha\beta$ in $R$, with $\alpha,\beta\notin k$. Write $\alpha=\sum \alpha\_i$ and $\beta=\sum \beta\_j$, where the $\alpha\_i$ and $\beta\_j$ are homogeneous (of the appropriate degrees). [Note: We implicitly only consider terms with nonzero support in the remainder.]


We can write $\alpha\_i=c^{e\_i}\alpha\_i'$ with $e\_i$ maximal. Similarly write $\beta\_j=c^{e\_j'}\beta\_j'$. Fix $m\_1$ maximal with $\deg(\alpha\_{m\_1}')$ maximal among $\{\deg(\alpha\_i')\}$. Similarly, fix $m\_2$ maximal with $\deg(\beta\_{m\_2}')$ maximized. 


Case 1: $\deg(\alpha\_{m\_1}')=0$ and $\deg(\beta\_{m\_2}')=0$. In this case $\alpha$ and $\beta$ are polynomials in $c$, which contradicts the irreducibility of $f$.


Case 2: Without loss of generality, $\deg(\alpha\_{m\_1}')>0$.


Consider the degree $m\_1+m\_2$ coefficient of $f(c)=\alpha\beta$. On the one hand, since $f(c)$ is a polynomial in $c$, it is either zero or a $k$-multiple of a power of $c$, say $c^{e}$. On the other hand, this term on the right-hand side is


$$c^{e\_{m\_1}+e\_{m\_2}'}\alpha\_{m\_1}'\beta\_{m\_2}' + \sum\_{(i,j)\neq (m\_1,m\_2)\ :\ i+j=m\_1+m\_2}\alpha\_i\beta\_j.$$


Each term in the big sum is divisible by strictly *more* powers of $c$ than $e\_{m\_1}+e\_{m\_2}'$, by maximality of degrees. Also $e>e\_{m\_1}+e\_{m\_2}'$, since $\deg(\alpha\_{m\_1}')>0$. This gives us the necessary contradiction. 


