Polynomials $X^n$ and $(1-X)^n$ are coprime

Question

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.

Answer 1

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.

Answer 2

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 $X$, $\gcd(X^n, (1-X)^n) = 1$.

Answer 3

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.

Answer 4

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 $\forall(n,m)\in\mathbb{N}^2,\,\gcd(P^n,Q^m)=1$).

Answer 5

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.