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This particular question is a part of an assignment which could not be discussed due to pandemic .

Question (a) Consider the ring R of polynomials in n variables with integer coefficients . Prove that the polynomial $f(x_1 , x_2 ,\ldots, x_n) = x_1 x_2\cdots x_n $ has $2^{n+1}-2$ non-constant polynomials in R dividing it .

Attempt: (a) any polynomial in R dividing f would be of form ${x_{1}}^{i} ...{x_{n}}^{i}$ i=0 or 1. So , there will be $2^{n}-1$ polynomials . But the answer is 2 times my answer.

Thanks!!

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    $\begingroup$ Hint for (a): does $-x$ divide $xy$? $\endgroup$
    – Arturo Magidin
    Sep 2, 2020 at 17:30
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    $\begingroup$ And also after you split, you should be easily able to make titles that aren't useless. $\endgroup$
    – rschwieb
    Sep 2, 2020 at 18:04

2 Answers 2

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Note that if $x_i ...x_k $ divides f then - ($x_i ...x_k$) also divides f . So, correct answer is $2^{n+1}-2$.

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In a) you forgot the coefficient: $-x_1$ divides $x_1x_2$. This gives $2^{n+1}-2$ divisors because the coefficient can be $1$ or $-1$.

b) The phrase "So, the elemets of order 11 is fixed" does not make sense.

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  • $\begingroup$ I meant that number of elements of order 11 is always 10 . Is it fine now ? $\endgroup$
    – user775699
    Sep 2, 2020 at 17:40
  • $\begingroup$ The phrase in the OP is the same. Your question is "Am I correct ?". The answer is "no". $\endgroup$
    – markvs
    Sep 2, 2020 at 17:50

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