$\mathbb{Q}[x]/ \langle α(x) \rangle$ if the polynomial is reducible

Question

I have some questions in the ideals of quotient rings. I know that there is some similar posts, but i would like to ask you some more general than the previous posts questions.

If we have the quotient ring $\mathbb{Q}[x]/ \langleα(x) \rangle$. What can we conclude in the following cases:

1. If $α(x) $ is an irreducible polynomial in $\mathbb{Q}[x]$ $\iff \mathbb{Q}[x]/ \langleα(x) \rangle$ is a field, and the only ideals of this ring are$\langle 0_{\mathbb{Q}[x]/ \langleα(x) \rangle} \rangle = \{\langleα(x) \rangle \}$ and $\mathbb{Q}[x]/ \langleα(x) \rangle$.
2. But what happens if $α(x)$ is reducible? And how can we find all the ideals of $\mathbb{Q}[x]/ \langleα(x) \rangle$?
3. $\mathbb{Q}\ \text{ is a field} \implies\mathbb{Q}[x]\ \text{is PID}$. Can we claim, from this fact, that $\mathbb{Q}[x]/ \langle α(x) \rangle$ is a PID?

Thank you.

Answer 1

1. You should know that $\mathbb Q[x]$ is an Euclidean domain, that is, given any polynomial $\alpha,\beta \in \mathbb Q[x]$, there exists $q,r \in \mathbb Q[x]$ with $\deg r < \deg \beta$ (set $\deg 0 = - \infty$ or add the case $r=0$ on the side, as you wish) such that $\alpha = q \beta + r$. (The degree serves as an Euclidean norm.) In particular, $\mathbb Q[x]$ is a PID, hence a principal ideal is prime if and only if a principal generator for it is irreducible. You should also be able to prove that in a PID, prime ideals are maximal ; this is a simple "suppose it is not maximal, pick an ideal above it and find the contradiction" argument. Therefore an ideal generated by an irreducible polynomial is maximal, i.e. the quotient is a field. As for the converse, if the quotient is a field, the ideal is maximal, thus prime, thus the generator $\alpha$ is irreducible.

2. If $\alpha$ is reducible, you can show that $\mathbb Q[x]$ is a UFD (because it is a PID), so we can factor $\alpha = \beta_1^{b_1} \cdots \beta_n^{b_n}$ where each $\beta_i$ is an irreducible polynomial (and the $\beta_i$'s are pairwise distinct). From the Chinese Remainder Theorem, we have an isomorphism of rings $$ \mathbb Q[x]/\langle \alpha \rangle \simeq \prod_{i=1}^n \mathbb Q[x]/\langle \beta_i^{b_i} \rangle. $$ The ideal structure becomes a little bit more complicated for this ring. We can discuss it in the comments if you wish.

3. Not if $\alpha$ is not irreducible, because remember, a PID is an integral domain. If $\alpha$ were reducible, say $\alpha = \beta \gamma$ with $\deg \beta, \deg \gamma < \deg \alpha$, then $(\beta + \langle \alpha \rangle)(\gamma + \langle \alpha \rangle) = 0$ in $\mathbb Q[x]/\langle \alpha \rangle$. But this is essentially the only constraint. Note that if $\alpha \neq 0$ and $\alpha$ is irreducible, $\mathbb Q[x]/\langle \alpha \rangle$ is a field ; this follows for many reasons, one of them being the following : since $\langle \alpha \rangle$ is a non-zero prime ideal in the PID $\mathbb Q[x]$, it is maximal in $\mathbb Q[x]$, so $\langle 0 \rangle$ is maximal in $\mathbb Q[x]/\langle \alpha \rangle$ by the prime ideal correspondence.

Hope that helps,

Answer 2

• If $\alpha(x)$ is reducible, if factors as a product of irreducible polynomials: $$\alpha(x)=p_1(x)^{r_1}\dotsm p_n(x)^{r_n},$$ you can apply the Chinese remainder theorem: $$\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)\simeq \mathbf Q[x]/\bigl(p_1(x)^{r_1}\bigr)\times\dots\times\mathbf Q[x]/\bigl(p_n(x)^{r_n}\bigr) $$ So in this case, $\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)$ is never an integral domain. More precisely:
• either $\alpha(x)$ has a multiple factor ($r_i>1$) and $\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)$ has a factor $\mathbf Q[x]/\bigl(p_i(x)^{r_i}$ which is a local ring, with a nilpotent maximal ideal $\bigl(p_i(x)\bigr)/\bigl(p_i(x)^{r_i}\bigr)$
• or $\alpha(x)$ has no multiple factor, $\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)$ is the product of fields $$\mathbf Q[x]/\bigl(p_1(x)\bigr)\times\dots\times\mathbf Q[x]/\bigl(p_n(x)\bigr),$$ which is a reduced ring (no non-zero nilpotent elements).

3: $\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)$ cannot be a P.I.D. if $\alpha(x)$ is reducible, since it's not even a domain. However, if $\alpha(x)= p(x)^r$ (only one irreducible factor), $\mathbf Q[x]\big/\bigl(\alpha(x)\bigr)$ its ideals are indeed principal – actually they're generated by the $p(x)^k,\enspace 0\le k\le r$.