Prove or disprove $R= \mathbb Q[x]/\langle x^3-x^2+x-1 \rangle$ is an integral domain.

Question

I've got $R$ is not a field since the polynomial is reducible in $ \mathbb Q[x]$. Is it possible to say anything from this?

Answer 1

Recall that $A/I$ is an integral domain if and only if $I$ is a prime ideal.

Further, $$x^3-x^2+x-1=(x^2+1)(x-1).$$

Do you see why this means the ideal generated by $x^3-x^2+x-1$ can't be prime? Further, do you see how this allows you to find elements $a$ and $b$ of the quotient ring such that $a,b\neq 0$ but $ab=0$?

Answer 2

We can factor $f(X)=X^3-X^2+X-1=(X-1)(X^2+1)\in \mathbb{Q}[X]$. Since $(X-1,X^2+1)=1$, we have that $P(X)=X-1$ and $Q(X)=X^2+1$ generate comaximal ideals.

By the Chinese remainder theorem, $\mathbb{Q}[X]/((X-1)(X^2+1))\cong \mathbb{Q}[X]/(X-1)\times \mathbb{Q}[X]/(X^2+1)\cong \mathbb{Q}\times\mathbb{Q}[i]$. The latter of which is not an integral domain as $(0,1)(1,0)=(0,0)$.

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Further, this produces elements $a,b$ of $\mathbb{Q}[X]/(X^3-X^2+X-1)$ such that $a\neq 0$ and $b\neq 0$ but $ab=0$. Let $\pi:\mathbb{Q}[X]\rightarrow \mathbb{Q}[X]/(X^3-X^2+X-1)$ be the canonical projection with $\pi(X)=x$. Then every element of the quotient can be written as a polynomial in $x$ with degree less than 3 (this is provable by Euclidean division). Now $a=x-1$ and $b=x^2+1$ are in the quotient ring but $ab=(x-1)(x^2+1)=x^3-x^2+x-1\equiv 0 \pmod{f(X)}$.