Prove that the following polynomial is irreducible in $ \mathbb Z[x,y]$: \[ x^{200}y^5+x^{51}y^{100}+x^{106}-4x^{100}y^5+x^{100}-2y^{100}-2x^6+4y^5-2\]
Problem
Source: Iranian National Olympiad (3rd Round) 2008
Tags: algebra, polynomial, algebra proposed
sumita
30.08.2008 22:52
we use generalized Esenestein .(the proof is the same as simple one) $ f(x,y) = y^{100}(x^{51} - 2) + y^5 \times {(x^{100} - 2)}^2 + (x^6 - 2) \times (x^{100} - 2)$ but we know that $ x^{100} - 2$ is a prime in ring of polynomial so applying isenestain will get to the point.because $ x^{100} - 2$ divide all of coeficients of polynomial(related to $ y$)expect$ x^{51} - 2$. and also $ (x^{100} - 2)^2$dosent divide $ (x^6 - 2) \times (x^{100} - 2)$
ZetaX
31.08.2008 01:32
isenestain -> Eisenstein
Rofler
31.08.2008 02:59
Extended eisensteins wont work, since $ (x^{100} - 2)^2$ doesn't divide $ (x^6 - 2)(x^{100} - 2)$
sumita
31.08.2008 03:06
the condition of eisenstein is that $ (x^{100}-2)^2$ shouldnt divide $ (x^6-2)(x^{100}-2)$
Gaussian_cyber
29.07.2020 21:14
$P(x,y) = y^{100}(x^{51} - 2) + y^5 \times {(x^{100} - 2)}^2 + (x^6 +1) \times (x^{100} - 2)$
FTSOC $P(x,y)=A(x,y)B(x,y)$ such that $A,B$ nonconstant
Now several claims east to check
Claim 1: for every $z \in \mathbb{Z}$ we have $ gcd(z^{100} -2 , z^6 +1) =1$
Claim 2: for every odd $z \in \mathbb {Z} $ we have $gcd(z^{100}-2 , z^{51} -2)=1$
Claim 3: $gcd(x^{100} -2, x^{51} -2) =1$
Now fix $y$ and vary $x$. notice that for every $x \in \mathbb{Z}$ we have $ x^{100} -2$ has an odd prime divisor with odd degree.
So by Eisenstein we have for every integer x the $ P(x,y)$ is irreducible. so one of $A , B$ becomes constant. Notice this can't happen that one of $A,B$ has $y$ terms in it and becomes constant for every integer $x$. So suppose wlog $B$ does not have $y$ term in it.
so $P(x,y)=A(x,y).B(x)$ So consider $y$ terms in $P$. Their $ x$ coefficiant should be divisible by $B(x)$. But by claims we have $gcd(x^{100} -2,x^{51} -2)=1$ absurd $\blacksquare$