Square roots of polynomials are linearly independent I would like to know how to prove the following (or maybe it's not true?)
Assume we have a set of squarefree,monic,distinct polynomials $P_1(x),...P_n(x)$ over $R$, so that they are all positive for $x>=0$.
Then I want to show the following: for every nonzero linear combination of $\sqrt P_1(x),..,\sqrt P_n(x)$ over $R$ and for every interval to the right of $0$, $(a,b)$, the linear combination is non constant on $(a,b)$.
I have tried taking derivatives, but seems impossible to do this :(. This is not a hw, but a result that I often myself in need of.
 A: This is not quite true as stated, since we can take $P_1(x)=1$, so $\sqrt{P_1(x)}=1$ is constant.
We can show that if $\sum_i c_i\sqrt{P_i(x)}=0$ on $(a,b)$, then $c_i=0$ for each $i$. More generally we will show by induction on $n$ that if
$$
  \sum_{i=1}^n\frac{f_i(x)}{P_i(x)^k}\sqrt{P_i(x)}=0
$$
on $(a,b)$, where the $f_i(x)$ are polynomials and $k$ is a nonnegative integer, then $f_i(x)$ is identically zero for each $i$. The case $n=1$ is clear.
Suppose the statement holds for $n-1$. If all the $f_i(x)$ are identically zero we are done, so suppose wlog $f_1(x)$ is not zero. By shrinking the interval, we may suppose $f_1(x)\neq0$ for $x\in(a,b)$. Let
$$
  g_i(x)=\frac{f_i(x)}{P_i(x)^k}\sqrt{P_i(x)}.
$$
Then $\sum_ig_i(x)=0$. Dividing by $g_1(x)$ and differentiating,
$$
  \sum_{i=2}^n\frac{d}{dx}\frac{g_i(x)}{g_1(x)}=0.
$$
We have
$$
  \frac{d}{dx}\frac{g_i(x)}{g_1(x)}=\frac{P_i(x)^{1/2-k}f_i'(x)}{g_1(x)}-\frac{g_i(x)}{g_1(x)}\left(\frac{f_1'(x)}{f_1(x)}+(k-1/2)\left(\frac{P_i'(x)}{P_i(x)}-\frac{P_1'(x)}{P_1(x)}\right)\right).
$$
Thus
$$
  f_1(x)P_1(x)g_1(x)\frac{d}{dx}\frac{g_i(x)}{g_1(x)}=\frac{h_i(x)}{P_i(x)^{k+1}}\sqrt{P_i(x)}
$$
for some polynomial $h_i(x)$. Thus
$$
  \sum_{i=2}^n\frac{h_i(x)}{P_i(x)^{k+1}}\sqrt{P_i(x)}=0.
$$
By induction, $h_i(x)$ is identically zero for $i=2,\ldots,n$, so $g_i(x)=k_ig_1(x)$ for some constant $k_i$. Squaring,
$$
  f_i(x)^2P_1(x)^{2k-1}=k_i^2f_1(x)^2P_i(x)^{2k-1}.
$$
Since the $P_i$ are square free and distinct, comparing irreducible factors we see that $f_i$ must be zero for $i>1$. This implies $f_1$ is also zero, completing the induction.
