Exercise 10.15

Proof. Write $f(x_0,\dots ,x_n)=x_0^{n+1}+p{x}_1^{n+1}+p^2{x}_2^{n+1}+\cdots +p^n{x}_n^{n+1}$.

Reasoning by contradiction, suppose that $\bar{a}=[{\alpha }_0,\dots ,{\alpha }_n]$ is a zero of $f$, where ${\alpha }_i\in \mathbb{Q}$ for $i=0,\dots ,n$. Using a common denominator $c$ for these rational numbers, we can write

where $d=b_0\wedge \cdots \wedge b_n$ is the gcd of the $b_i$, so that the $a_i\in \mathbb{Z}$ satisfy $a_0\wedge \cdots \wedge a_n=1$.

Then

where the integers $a_i$ are not all divisible by $p$.

To obtain a contradiction, we will show that all the $a_i$ are divisible by $p$.

$p\mid -p{a}_1^{n+1}-p^2{a}_2^{n+1}+\cdots -p^n{a}_n^{n+1}=a_0^{n+1}$, thus $p\mid a_0$.

Reasoning by induction, suppose that $p$ divides $a_0,\dots ,a_k$, where $k<n$. Then $p^{n+1}\mid a_0^{n+1}+p{a}_1^{n+1}+p^2{a}_2^{n+1}+\cdots +p^k{a}_k^{n+1}$, therefore

Since $n>k$, $p\mid a_{k+1}^{n+1}+p{a}_{k+2}^{n+1}+\cdots +p^{n-k-1}{a}_n^{n+1}$, therefore $p\mid a_{k+1}^{n+1}$, thus $p\mid a_{k+1}$.

The induction is done. This proves that $p\mid a_0,\dots ,p\mid a_n$. This is a contradiction, since the $a_i$ are not all divisible by $p$. So the form $x_0^{n+1}+p{x}_1^{n+1}+p^2{x}_2^{n+1}+\cdots +p^n{x}_n^{n+1}$ has no zeros in $P_n(\mathbb{Q})$. □