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So I know that for it to be an integral domain it has to have the following properties:

  1. Commutative
  2. Has multiplicative identity
  3. No Zero-Divisors

and if $p(x)$ is reducible it can be written as $p(x) = (x-a)q(x)$ but I can't seem to link it together to prove that it violates one of the three properties.

Torched90
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  • Why should $p(x)$ have a linear factor? – Bernard Apr 07 '16 at 20:49
  • If $p(x)=q(x)h(x)$, then try showing that $q(x)$ and $h(x)$ are zero divisors by multiplying them together in the $F[x]/(p(x))$ ring. – Noble Mushtak Apr 07 '16 at 20:49
  • @NobleMushtak so if $q(x)h(x) = 0$ then one of them is a zero divisor if they are both non-zero polynomials, but how do I prove that they are equal to zero? – Torched90 Apr 07 '16 at 21:00
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    Well, you know that in $F[x]$, $q(x)h(x)=p(x)$. What is $p(x)$ equal to $F[x]/(p(x))$? – Noble Mushtak Apr 07 '16 at 21:05
  • @NobleMushtak I'm not sure what conclusion to make. Is it equal to zero? – Torched90 Apr 07 '16 at 21:11
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    You can try the contrapositive also by using the characterization of R/A is an integral domain if and only if A is a prime ideal and using the PIDness of $F[x]$ you can conclude that things that are prime are irreducible things. – Daniel Akech Thiong Apr 07 '16 at 23:00
  • As an addendum to the first comment, $x^4 + 4x^2 + 4 \in \mathbb{Q}[X]$ is reducible as $(x^2 + 2)^2$, but it is not divisible by a linear factor, i.e. has no roots in $\mathbb{Q}$. – D_S Apr 08 '16 at 22:22

2 Answers2

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Let $R = F[x]$. The important thing is that $R$ is a principal ideal domain: thus every nonzero nonunit factors as a product of irreducible elements. Every ideal of $R$ is generated by a single element in $R$, and that ideal is a prime ideal if and only if the element generating it is either $0$ or an irreducible element.

If $p$ is reducible, then the ideal $(p)$ generated by $p$ is not a prime ideal. Hence the quotient ring $F[X]/(p)$ is not an integral domain.

D_S
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Let's say $p(x)$ is reducible so that $q(x)h(x)=p(x)$ for non-constant polynomials $q(x)$ and $h(x)$.

Now, consider $q(x)h(x)+\langle p(x)\rangle$ in the ring $F[X]/\langle p(x)\rangle$. Normally, $q(x)h(x)=p(x)=p(x)$, so clearly, $q(x)p(x) \in \langle p(x)\rangle$. Therefore: $$q(x)h(x)+\langle p(x)\rangle=0+\langle p(x)\rangle$$

This means $(q(x)+\langle p(x)\rangle)(h(x)+\langle p(x)\rangle)=0+\langle p(x)\rangle$. Now, we need to prove that $q(x)+\langle p(x)\rangle$ and $h(x)+\langle p(x)\rangle$ are non-zero. Well, what does $q(x)+\langle p(x)\rangle$ being $0$ really mean? It means $q(x) \in \langle p(x)\rangle$, which means $q(x)=p(x)r(x)$ for some polynomial $r(x)$. However, $p(x)=q(x)h(x)$, so we have the following: $$q(x)=q(x)h(x)r(x)$$ This clearly implies that $h(x)r(x)=1$. However, this can only happen if $h(x)$ is a constant polynomial, which we assumed it wasn't. Therefore, $q(x)+\langle p(x)\rangle$ must be non-zero and $h(x)+\langle p(x)\rangle$ is also non-zero.

Thus, we have found two non-zero elements which, when multiplied, yield zero, which means that the ring $F[X]/\langle p(x)\rangle$ has zero divisors, meaning it is not an integral domain.

Noble Mushtak
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