# Prove that $q(x)$ does not divide $p_k(x)$

Let $n \in \mathbb{N}$ and let $p_1(x), p_2(x), ... *p_n(x)$ be $n$ irreducible polynomials over $\mathbb{R}$. Define the polynomial $p(x) = p_1(x) * p_2(x) *... *p_n(x) + 1$ where 1 is the constant polynomial $0x + 1$. Let $q(x)$ be an irreducible polynomial and suppose that $q(x)$ | $p(x)$. Prove that $q(x$) does not divide $p_k(x)$ for all $k \in${$1..n$}

My attempt so far:

If $q(x)$ is an irreducible polynomial and $q(x)$ | $p(x)$, then gcd[$p(x), q(x)] = q(x)$

So $ap(x) + bq(x) = 1$ for some $a,b$ $\in \mathbb{R}[x]$

So $a(p_1(x) * p_2(x) *..*p_k(x)*.. *p_n(x) + 1) + bq(x) = 1$

So $(a(p_1(x) * p_2(x) *... *p_n(x) + 1)p_k(x) + bq(x) = 1$

Therefore $cp_k(x) + bq(x) = 1$ (where $c = a(p_1(x) * p_2(x) *... *p_n(x) + 1)$

Hence $p_k(x)$ and $q(x)$ are relatively prime so $q(x$) does not divide $p_k(x)$ for all $k \in${$1..n$}

## 2 Answers

The only units in this ring are the nonzero real numbers. Suppose $q$ divides some $p_k$ and $q$ divides p. Then $q$ divides $1$ which is a contradiction because irreducible elements are not units.

If $q$ divides $p_k$, then $q$ divides $p_1 \cdots p_n = p - 1$. By hypothesis, $q$ divides $p$. Hence $q$ divides $p - (p-1) = 1$.