2
$\begingroup$

Let $n\in \mathbb N-\{0\}$ . How can we show that the polynomials $X^n$ and $(1-X)^n$ are coprime? Do we have to do an induction on $n$? it is clear that $X$ and $(1-X)$ are coprime by Euclid's algorithm. We can write $(1-X)^{n+1}=\sum_{k=0}^{n+1}{(-1)^k{k\choose n+1}X^k}$ but then how to proceed for the induction. thank you for your help.

$\endgroup$
1
$\begingroup$

We have according to Newton's identity:

$(1-X)^n=\sum\limits_{k=0}^n{n\choose k}(-1)^kX^{n-k}=1+X\sum\limits_{k=1}^n{n\choose k}(-1)^kX^{n-k-1}$

Put $P=\sum\limits_{k=1}^n{n\choose k}(-1)^kX^{n-k-1}$

It means that $\exists P\in\mathbb{C}[X],\,(1-X)^n-P(X)X=1$.

According to Bezout's theorem, $\gcd\left((1-X)^n,X\right)=1$ and so $\gcd\left((1-X)^n,X^n\right)=1$ (because if $\gcd(P,Q)=1$ then $\forall(n,m)\in\mathbb{N}^2,\,\gcd(P^n,Q^m)=1$).

$\endgroup$
4
$\begingroup$

The only irreducible divisor of $x^n$ is $x$. By the binomial theorem,

$$(1-x)^n=1+x\left(\sum_{k=1}^n(-1)^k{n\choose k}x^{k-1}\right).$$

But then if $x|(1-x)^n$ we have that $$x\bigg|\left((1-x)^n-x\left(\sum_{k=1}^n(-1)^k{n\choose k}x^{k-1}\right)\right).$$

Of course this means $x|1$ which is impossible. Hence no primes are shared by $x^n$ and $(1-x)^n$ so they are coprime.

$\endgroup$
  • $\begingroup$ Hugues : What do you mean by prime divisor here? irreducible ? $\endgroup$ – palio Jan 7 '15 at 7:30
  • $\begingroup$ @palio yes, irreducible is the same thing in this context. $\endgroup$ – Adam Hughes Jan 7 '15 at 7:35
2
$\begingroup$

Let $Y = 1 - X$, we have $$1 = (X + Y)^{2n-1} = \left( \sum_{k=0}^{n-1} + \sum_{k=n}^{2n-1} \right) \binom{2n-1}{k} X^k Y^{2n-1-k}$$ Let $j = 2n-1-k$ in the second sum and notice $\displaystyle\;\binom{n}{k} = \binom{n}{j}$, we get $$1 = \left( \sum_{k=0}^{n-1} \binom{2n-1}{k} X^k Y^{n-1-k} \right) Y^n + X^n \left( \sum_{j=0}^{n-1} \binom{2n-1}{j} X^{n-1-j} Y^j\right) $$ Since $1$ can be expressed as a linear combination of $X^n$ and $Y^n = (1-X)^n$ over the polynomial ring of $X$, $\gcd(X^n, (1-X)^n) = 1$.

$\endgroup$
1
$\begingroup$

Suppose $p(X)$ is a monic divisor of both $X^n$ and $(1-X)^n$. Then for $p(X) \mid X^n$, any root of $p(X)$ has to be $0$. Because of $p(X) \mid (1-X)^n$, any root of $p(X)$ has to be $1$. Therefore $p$ cannot have any roots and we must have $p(X) \equiv 1$, which is to say that $X^n$ and $(1-X)^n$ are coprime.

$\endgroup$
  • $\begingroup$ There are polynomials without roots in given fields, like $x^2+1$ over the reals, you need to account for algebraic closure considerations with this approach, or classify the divisors (it's definitely easily salvaged) $\endgroup$ – Adam Hughes Jan 7 '15 at 7:18
0
$\begingroup$

As you say, the case $n=1$ is clear. So suppose that $n >1$ and we know the result for $n-1$.

Suppose that the gcd is not $1$. Then some irreducible polynomial $f(X)$ of positive degree divides both $X^{n}$ and $(1-X)^{n}.$

Then (from the first, and using $f(X)$ irreducible) $f(X)$ divides $X$ or $X^{n-1}$ ( so, in any case, as $n>1$, $f(X)$ divides $X^{n-1}$). Simillarly, from the second, $f(X)$ divides $(1-X)$ or $(1-X)^{n-1},$ so in any case, as $n >1,$ $f(X)$ divides $(1-X)^{n-1}$. Hence we know that $f(X)$ divides both $X^{n-1}$ and $(1-X)^{n-1}$, so divides their gcd, which is 1 by induction, a contradiction as $f(X)$ has positive degree.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.