1

Let the real coefficient polynomials $$f(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$$ $$g(x)=b_{m}x^m+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}$$ where $a_{n}b_{m}\neq 0,n\ge 1,m\ge 1$, and let $$g_{t}(x)=b_{m}x^m+(b_{m-1}+t)x^{m-1}+\cdots+(b_{1}+t^{m-1})x+(b_{0}+t^m).$$ Show that

there exist positive $\delta$, such for any $t$ such that $0<|t|<\delta$, and such $f(x)$ and $g_{t}(x)$ be relatively prime.

I fell this result is very well, because although two polynomial are not relatively prime, we can do it to one of the polynomial tiny perturbation makes relatively prime.

But I can't prove my problem.

Thank you very much

Davide Giraudo
  • 172,925
math110
  • 93,304
  • If you can show that there are only finitely many $t$ for which they aren't relatively prime, you'll practically be done (I don't know that this is true, but I feel that that's the way to go). – Arthur Mar 04 '14 at 10:19
  • I think so,@Arthur,Thank you,But How prove it? Thank you – math110 Mar 04 '14 at 10:20
  • Seems obvious (haven't worked through the steps). $f(x) $ has finitely many roots that $g_t (x)$ wants to avoid. – Calvin Lin Mar 04 '14 at 10:23

3 Answers3

2

Here is a series of steps, which seem mostly true to me.

Fact: $f(x)$ has at most $n$ distinct roots.

Claim: There exists a map $G: [0,1] \rightarrow (\alpha_1, \alpha_2, \ldots, \alpha_m)$ which is differentiable in each coordinate, and $ \alpha_i$ are roots of $ g_t(x)$ (with multiplicity).

Possible Proof: Use Inverse/Implicit function theorem till 2 roots meet. Then be very careful?

Claim: On any open interval, $\frac{dG}{d\alpha_i}$ is not identically 0.

Proof: If it does, then there is a number $ \alpha$ such that $ g_t(\alpha) = 0 $ on that open interval. But this implies that $ \sum t^i \alpha^{m-i} = 0 $ infinitely often, which is a contradiction (since there is at most $m$ values of $t$ that can satisfy the polynomial).

Hence, the result follows, where we take a small enough $t$, such that the roots of $g_t$ all avoid the roots of $f$. The final claim allows us to move off of common roots of $f$ and $g_0$ .

Calvin Lin
  • 68,864
  • Nice! if you can post your full solution!+1 – math110 Mar 04 '14 at 10:45
  • @math110 I'm stuck at the first claim. I don't remember enough of my analysis for a proper proof of it, but I believe that it is (very likely) true. Given that, the above is a solution. – Calvin Lin Mar 04 '14 at 10:47
2

let $x_i$ be the roots of $f$.

$g_t$ and $f$ are not primes iif $g_t(x_i) \neq 0$ for every $x_i$.

So $g_t$ and $f$ are primes iif $t$ no in the finite set of roots of $$b_{m}x_i^m+(b_{m-1}+t)x_i^{m-1}+\cdots+(b_{1}+t^{m-1})x_i+(b_{0}+t^m)$$

Sylvain Biehler
  • 478
  • You still require slightly more work, to establish that roots of $g_t$ and $f$ do not intersect for a small neighbor hood of $t$, even though $g_0$ and $f$ may not be relatively prime. – Calvin Lin Mar 04 '14 at 10:36
  • since there is a finite set of forbidden values ${t_i}$, I claim it exists $\delta$, such no forbidden value verifies $0<|t_i|<\delta$ – Sylvain Biehler Mar 04 '14 at 10:51
  • Ah, I see what you are doing now. This looks good (+1). My comment was prejudiced by my approach, and doesn't apply to yours. – Calvin Lin Mar 04 '14 at 10:57
0

You just need to check that the resultant $r(t)=Res_x(f(x), g_t(x))$, which is a polynomial in $t$, is not the zero polynomial. Any root of $r(t)$ marks a value of $t$ where both polynomials have a common factor. If $r(t)\ne 0$, the polynomials $f(x)$ and $g_t(X)$ are relatively prime.

And as in the other answers, as $r(t)$ only has a finite number of roots...

Lutz Lehmann
  • 126,666