Find all $\mathbb{Q}$-linear maps $\phi : \mathbb{Q}[x]\to\mathbb{Q}[x]$ that send irreducible polynomials to irreducible polynomials 
Let $\mathbb{Q}[x]$ denote the vector space over $\mathbb{Q}$ of polynomials with rational coefficients in $x$. Find all $\mathbb{Q}$-linear maps $\phi : \mathbb{Q}[x]\to\mathbb{Q}[x]$ that send irreducible polynomials to irreducible polynomials.

If $\mathbb{K}$ is a field over which a vector space $V$  is defined and $W$ is a vector space, a $\mathbb{K}$-linear map $T :V \to W$ satisfies $T(cx+y) = cT(x)+T(y).$ In particular, for any such linear map $T, T(0) = T(0)+T(0)\Rightarrow T(0)=0$. Also, linear maps are uniquely determined by their values on a countable or finite basis, since if $\{a_1,a_2,\cdots\}$ is a basis for $V$ and $T:V\to W$ is a linear map, for any $x\in V, T(x) = \sum_{i=1}^k b_i T(a_{j_i})$ where $x = \sum_{i=1}^k b_i a_{j_i}$ is the unique representation of $x$ as a linear combination of basis vectors. In particular, $\{1,x,\cdots\}$ is a countable basis for $\mathbb{Q}[x]$ so it suffices to find the values of $\phi(x^i)$ for all $i\ge 0$.
Note that in $\mathbb{C}[x]$, the only irreducible polynomials are the polynomials of degree $1$ since it is known that every polynomial of degree $n\ge 1$ has a complex root. In $\mathbb{R}[x]$, every odd degree polynomial has a real root since complex roots come in pairs. Hence no odd degree polynomial with degree at least 3 is irreducible. Note that the set of units in $\mathbb{K}[x]$ for a field $\mathbb{K}$ is precisely $\mathbb{K}\backslash \{0\}$, since for any two elements $f,g \in R[x]$ where R is a ring and $f$ and $g$ have leading coefficients that are units, $\deg(fg) = \deg(f) + \deg(g)$. In general for any ring $R$ and $f,g\in R[x], \deg(fg)\leq \deg(f)+\deg(g)\tag{1}$. In $\mathbb{Q}[x]$, any polynomial of degree $1$ is irreducible and the polynomials of degree $2$ that are irreducible are precisely those without rational roots. Also, by Gauss' lemma, in a unique factorization domain (UFD) $R$, if $\mathbb{K}$ is the field of fractions of $R$ then if $f \in R[x]$ equals $gh$ for some $g,h \in \mathbb{K}[x],$ there exists a unit $u \in \mathbb{K}$ so that $ug, u^{-1} h \in R[x].$ In particular for the UFD $\mathbb{Z}$, its field of fractions is $\mathbb{Q}$ so if a polynomial with integer coefficients can be written as a product of two nonconstant polynomials with rational coefficients, then it can also be written as a product of two nonconstant polynomials with integer coefficients. Also, if $f\in R$ is irreducible, then $f$ is also irreducible in $R[x]$ whenever $R$ is an integral domain as then (1) always holds with equality.
For the given problem, it might be the case that if $f + cg$ is irreducible (in $\mathbb{Q}[x]$) for all $c\in\mathbb{Q}$, then either $g=0$ and $f$ is irreducible or $f$ is of degree 1 and g is a constant or g is linear. If this claim holds, if $f,g$ are polynomials so that $f+cg$ is irreducible for all $c\in\mathbb{Q},$ then since $\phi(f+cg) = \phi(f)+c\phi(g),\phi(f) + c\phi(g)$ are irreducible for all $c\in\mathbb{Q}$. Hence $\phi(g) = 0$ or $\phi(f)$ has degree $1$ and $\phi(g)$ is a constant or linear.
As an attempt to prove the claim, I think one can proceed using a proof by contradiction. Suppose $f(x)$ has degree $1$ and $g$ is nonzero. Write $f(x)=dx+e, d,e\in\mathbb{Q}$. If $g$ has degree 2, say $g = ax^2+bx+c$ for some $a,b,c\in\mathbb{Q}$, then $f(x)+rg(x) = rax^2 + (rb+d)x+(rc+e)$. We need to choose $r$ so that this quadratic has a rational root, which will be the case provided its discriminant is the square of a rational number. But I'm not sure how to find such an $r$ or if it (always) exists.
The product of two irreducible polynomials is obviously reducible but the sum may or may not be irreducible. For instance, $x+x^2+x+1$ is reducible even though both $x$ and $x^2+x+1$ are irreducible over $\mathbb{Q}[x]$. Every polynomial in $\mathbb{Q}[x]$ or $\mathbb{R}[x]$ can be written (uniquely) as a product of irreducibles (note that every Euclidean domain is a PID and every PID is a UFD).
 A: $\newcommand\Q{\mathbb Q}$There are 5 families of $\Q$-linear maps $\phi : \Q[x]\to\Q[x]$ that send irreducible polynomials to irreducible polynomials. The summary is, because of the "rigid" and "distinguishing" structure of irreducible polynomials, all such maps are sort of trivial.

*

*The degree of $\phi(1)$ is at least $2$.
There exists an irreducible polynomial $q$ such that $\phi(1)=q$ and $\phi(x^i)=0$ for all $i\ge1$.

*The degree of $\phi(1)$ is $1$.
There exist $a,b,c,d\in\Q$, $a\not=0$, $ad\not=bc$ such that $\phi(1)= ax+b$, $\phi(x)=cx+d$ and $\phi(x^i)=0$ for all $i\ge2$. .

*The degree of $\phi(1)$ is $0$.
There exist $a, b, c\in\Q$, $a\not=0$, $b\not=0$ such that  $\phi(1)= a$, $\phi(x)=bx+c$.

*

*$\phi(x^i)=0$ for all $i\ge2$.

*$\phi(x^i)=0$ for all $i\ge2$ except that for one integer $j\ge2$ and $0\not=d\in\Q$ such that $\phi(x^j)=d(bx+c)$.

*$\phi(x^i)= a(\frac{bx+c}a)^i$ for all $i\ge2$.



A: If $f(x) + rg(x)$ has some zero then it has a factor. Thus any desired zero can be had with the appropriate choice of $r$ easily solved from $$ r = \frac{-f(x_0)}{g(x_0)} $$ for some $x_0$.
Choose $x_0 = 0$ and we get $$r = \frac{-e}{c}$$
Then $f(x) + rg(x)$ should have factor $x$:
$$
= f(x) - \frac{e}{c} g(x) $$
or equivalently $cf(x) - eg(x)$ should have a factor $x$:
$$
= cf(x) - eg(x)
$$
$$
= c(dx+e) - e(ax^2 + bx + c)
$$
$$
= cdx + ce - aex^2 - bex - ce
$$
$$
= -aex^2 + (cd - be)x + ce - ce
$$
$$
= -aex^2 + (cd - be)x
$$
$$
=x(-aex + cd - be)
$$
and we see that it has the factor $x$
