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

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.

## closed as off-topic by Thomas, Parcly Taxel, user26857, user91500, R_DSep 12 '16 at 4:55

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## 2 Answers

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,

• @Patrick Da Silva Thank you for your answer. I have some more questions. 1) If, for instance we have $\mathbb{Q}[x]/ \langle x^3 \rangle$ what we can conclude? 2) Can we give a more simple answer than the Chinese Remainder Theorem? – Chris Sep 8 '16 at 21:24
• @Chris : The Chinese Remainder Theorem gives you the "identity" isomorphism in this case, so it's not helpful. An ideal of this ring is contained in its only maximal ideal, namely $\langle x \rangle/\langle x^3 \rangle$. This ideal corresponds to a polynomial dividing $x^3$ and divisible by $x$, hence is equal to $x$, $x^2$ or $x^3$. I let you figure out the general case! – Patrick Da Silva Sep 8 '16 at 21:37
• @PatrickDaSilva Your answers are really helpful. But, why do we have $\langle x \rangle / \langle x^3 \rangle$? In the case we have the quotient $A/ \langle x^3 \rangle$ is required to have "$A=\langle f(x) \rangle \subset \mathbb{Q}[x]$"? Namely $A$ must be a principal ideal, and why? – Chris Sep 8 '16 at 21:46
• This is Krull's Theorem : any proper ideal is contained in a maximal ideal. By the prime ideal correspondence, the only maximal ideal of the quotient is the one generated by $x$. – Patrick Da Silva Sep 8 '16 at 23:25
• 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 $p(x)^k,\enspace 0\le k\le r$.