currently reading a textbook which states the following - if a polynomial $f$ is reducible in $\mathbb{Z}$, so long $n$ does not divide the highest coefficient of $f$ it is irreducible in $\mathbb{Z}_n$. don't they mean the opposite? if that is the case, why can $n$ not divide the highest coeff of $f$? is it because it will have a lower degree in $\mathbb{Z}_n$?
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It should say that if the polynomial is reducible over $\mathbb{Z}$ then it is reducible over $\mathbb{Z}_n$, if the lead coefficient is not divisible by $n$. The ordinary use of this result is to prove irreducibility over $\mathbb{Z}$ by showing irreducibility over a well-chosen $\mathbb{Z}_n$. Maybe that is the reason for the verbal slip.
The reason for the restriction is essentially the one you gave. If the lead coefficient is divisible by $n$, then a splitting of $P(x)$ as a product of polynomials of lower degree may not induce such a splitting when we reduce modulo $n$. As an extreme example, let $n=2$, and consider the polynomial $4x^2-1$.
André Nicolas
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ah, i see it now. leading coefficient divisible by p -> leading coefficient of one of the factors divisible by p -> one of the factors vanishes. thanks! – irreducibilityeisensteinwow Sep 19 '15 at 21:23
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I can arrange for a polynomial to be reducible over the integers, have lead coefficient divisible by $n$, and still reducible modulo $n$. But the point is that there are some polynomials which are reducible over the integers but not mod $n$. – André Nicolas Sep 19 '15 at 21:33
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For example let $P(x)=(3x^2-1)(x^2)$. Reducible over the integers. Let $n=3$. The resulting polynomial, which can be written as $-x^2$, is reducible mod $3$. But sometimes when the lead coefficient is divisible by $n$ we do not have reducibility mod $n$. – André Nicolas Sep 19 '15 at 21:39
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You are welcome. It can be puzzling when one meets a typo in a book. But in text of any length, typos are almost inevitable, and can elude even very diligent proofreading. – André Nicolas Sep 20 '15 at 14:40