Polynomial transformation of the roots of another irreducible polynomial.

Question

Suppose I have some monic irreducible polynomial $g(x)$ in $\mathbb{Z}[x]$ with distinct roots $r_1,r_2,\dots,r_n$. Suppose $f(x)$ is some other polynomial, not necessarily irreducible. Is there anything I could say about $f(r_1),f(r_2),\dots,f(r_n)$?

For example: let $g(x)=5-10x^3+5x^4+8x^5-4x^6-2x^7+x^8$ and $f(x)=1-3 x^2+x^4$. Here $g$ is irreducible but $f$ is not. In this case, if $r_1,r_2,\dots,r_8$ are the roots of $g$, then $f(r_1),f(r_2),\dots,f(r_8)$ are also algebraic integers of degree 8 and are all conjugate (as is they have the same minimal polynomial).

Are there conditions on $g$ and $f$ so that this will always happen?

Also, in the above example $\prod_{i=1}^8 r_i=5$ and $\prod_{i=1}^8 f(r_i)=2000$. Are there conditions on $f$ that make it so that $\prod r_i$ divides $\prod f(r_i)$?

Answer 1

Yes, $f(r_1),\ldots,f(r_n)$ are conjugate.

Generally, let $\mathbb{F}$ be a field, let $g\in\mathbb{F}[x]$ be irreducible, and denote the roots of $g$ by $r_1,\ldots,r_n$. Let $f\in\mathbb{F}[x]$ some other polynomial, and observe the set $\{f(r_1),\ldots,f(r_n)\}$. Assume that for some $h\in\mathbb{F}[x],\quad h(f(r_1))=0.$ Then $r_1$ is a root of the polynomial $h\circ f$, meaning that the latter is divisible by $g$ and for every $j=1,\ldots,n\quad h(f(r_j))=h\circ f(r_j)=0.$ Thus $f(r_1),\ldots,f(r_n)$ are the roots of the same polynomials, or in other words they have the same minimal polynomial.

Note, however, that $f(r_1),\ldots,f(r_n)$ need not be distinct. For example, if $g=x^4+1$ then the roots are $1,i,-1,-i$. If we take $f=x^2$ then of course $f(r_1)=f(r_3),f(r_2)=f(r_4)$.

Answer 2

Since $g$ is monic, the resultant of $f(z)$ and $g(z)$ is $res(f,g) = \prod_r f(r)$ where the product is over the roots $r$ of $g$, while of course $\prod_r r = \pm g(0)$.
This is true not only for roots in $\mathbb C$, but (since $res(f,g)$ is a polynomial expression in the coefficients of $f$ and $g$) over any splitting field of $g$.

In this case $g(0)=5$, and $g$ splits over ${\mathbb Z}_5$: $g(z) = z^5 (z+3)^2(z+2) \mod 5$, and we have $(f(0), f(-3), f(-2)) \equiv (1,0,0) \mod 5$. Since there is at least one $0$ there, we have $res(f,g) \equiv 0 \mod 5$.

Somewhat more generally, suppose $g$ is monic and $g(0) = \pm \prod_{i=1}^k p_k$ where $p_k$ are distinct primes (maybe someone else can do the case where $g(0)$ is not squarefree). In order for $res(f,g)$ to be divisible by $g(0)$, we need it to be divisible by each $p_i$, and what we need for that is that $f$ and $g$ have a root in common in some field extension of ${\mathbb Z}_{p_i}$.