Let \(f(x)\in\mathbb Q[x]\). Show that there exists a nonzero \(g(x)\in\mathbb Q[x]\) such that every nonzero term of \[ f(x)g(x) \] has prime exponent; equivalently, \[ f(x)g(x)=a_2x^2+a_3x^3+a_5x^5+\cdots+a_px^p \] for suitable coefficients and primes.
Solution
1. If \(f=0\), the assertion holds.
Take \[ g(x)=1. \] Then \[ f(x)g(x)=0, \] which has no nonzero terms, so the condition on nonzero exponents is vacuously satisfied.
2. If \(f\) is a nonzero constant, the assertion holds.
Write \[ f(x)=c \] with \(c\in\QQ^\times\), and take \[ g(x)=c^{-1}x^2. \] Then \[ f(x)g(x)=x^2, \] whose only exponent is the prime \(2\).
3. Suppose\[ d\coloneqq\deg f\geq1. \]Then the quotient\[ V\coloneqq\QQ[x]/(f) \]is a \(d\)-dimensional vector space over \(\QQ\).
Every polynomial has a unique remainder of degree less than \(d\) upon division by \(f\). Therefore the residue classes of \[ 1,x,\ldots,x^{d-1} \] form a basis of \(V\).
4. Choose distinct primes\[ p_0,p_1,\ldots,p_d. \]Then there are rational numbers\[ a_0,\ldots,a_d, \]not all zero, such that\[ \sum_{j=0}^{d}a_jx^{p_j} \equiv0 \pmod f. \]
3. They aretherefore linearly dependent, giving coefficients \(a_j\in\QQ\), not allzero, whose displayed linear combination represents the zero class.
5. The polynomial\[ h(x)\coloneqq \sum_{j=0}^{d}a_jx^{p_j} \]is nonzero and divisible by \(f\).
The exponents \(p_0,\ldots,p_d\) are distinct and the coefficients \(a_j\) are not all zero, so the monomials cannot cancel identically; hence \(h\neq0\). The congruence in step 4 says exactly that \[ f\mid h \] in \(\QQ[x]\).
6. There is a nonzero polynomial\[ g(x)\in\QQ[x] \]such that every nonzero term of \(f(x)g(x)\) has prime exponent.
By step 5, write \[ h(x)=f(x)g(x) \] for some \(g\in\QQ[x]\). Since both \(f\) and \(h\) are nonzero, \(g\neq0\). Moreover, \[ f(x)g(x) = h(x) = \sum_{j=0}^{d}a_jx^{p_j}, \] and every exponent \(p_j\) occurring in this expression is prime.
7. Q.E.D.
Steps 1 and 2 handle the cases \(\deg f\leq0\), while step 6 handles every nonconstant polynomial.