0

Let $\mathbb K$ be a field, and $f(x)=p(x)q(x),\,p(x),q(x)\in\mathbb K[x]$ where $p(x)$ and $q(x)$ are two distinct irreducible monic polynomials. Now, compute $\mathbb K[x]/(f(x))$, here the notation $(f(x))$ means the principle ideal generated by $f(x)$,

Here is my approach:
Notice $f(x)=p(x)q(x)$, we want to see what the relation between $(f(x))$ and $(p(x)),\, (q(x))$. It turns out that $(f(x))=(p(x))\cap(q(x))$. This is because for any element in $(f(x))$ it is "divisible" by $p(x)$ and $q(x)$ hence $(f(x))\subseteq(p(x))\cap(q(x))$. On the other hand, let $\rho(x)\in (p(x))\cap(q(x))$, then $\rho(x)=p(x)k(x)=q(x)l(x)$ for some $k(x)$ and $l(x)$ in $\mathbb K[x]$. Since $p(x)$ and $q(x)$ are distinct irreducible polynomials, this indicates $k(x)=q(x)k'(x)$ and $l(x)=p(x)l'(x)$ with $k'(x)$ and $l'(x)$ are two other polynomials in $\mathbb K$. Hence $\rho(x)$ is "divisible" by $f(x)$ i.e. $(p(x))\cap(q(x))\subseteq (f(x))\Rightarrow\, (f(x))=(p(x))\cap(q(x))$.

Then I don't know what to do next. I have a thought to show that $(p(x))$ and $(q(x))$ are comaximal but which may not be true.

Hamio Jiang
  • 527
  • The question is not well-posed without saying what you mean by "isomorphic twin" and "compute $\Bbb K[x]/f$" $\ \ \ $ – Bill Dubuque Jul 13 '17 at 14:56
  • @BillDubuque This is also my question... It is my assignment question without changing any words. It may ask you to write down all elements of the quotient ring or what it is isomorphic to...I don't know either... – Hamio Jiang Jul 13 '17 at 15:00
  • 1
    Do you know some form of CRT = Chinese Remainder Theorem? – Bill Dubuque Jul 13 '17 at 15:03
  • @BillDubuque Not yet. Maybe I could search it. Is it the division algorithm of polynomials on a field? – Hamio Jiang Jul 13 '17 at 15:04
  • No it's a theorem that generalizes $\mathbb{Z}/mn\mathbb{Z}\simeq \mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$ for $m\land n =1 $ – Maxime Ramzi Jul 13 '17 at 17:47

1 Answers1

1

Ideals $(p(x))$ and $(q(x))$ are indeed comaximal, for if not $$(p(x)),(q(x))\subset(p(x))+(q(x))\subset R$$ Which contradicts the maximality of both $(p(x))$ and $(q(x))$

\begin{align} \Bbb K[x]/(f(x)) &=\Bbb K[x]/(p(x)q(x)) \\ &=\Bbb K[x]/(p(x))(q(x)) \\ &=\Bbb K[x]/(p(x))\cap(q(x))\\ & \cong \Bbb K[x]/(p(x))\times \Bbb K[x]/(q(x)) \\ \end{align}

The first equality follows from $f(x)=p(x)q(x)$.

The second equality follows from Result-4

The third equality follows from Result-1,Result-2 and Result-3

The Isomorphism follows from The Chinese Remainder Theorem for Rings.

Result-1: In a PID the ideals generated by irreducibles are maximal.

Result-2: Let $R$ be a commutative ring with unity(CRU) and $I,J\subset R$ be two distinct maximal ideals in $R$. Then $I+J=R$

Result-3: Let $R$ be CRU and $I,J$ be two ideals of $R$ such that $I+J=R$. Then $IJ=I\cap J$

Result-4:Let $R$ be CRU. Then $(a)\cdot (b) = (a b)$ for any $a, b\in R$

Naive
  • 1,800