1
$\begingroup$

Find all values of $b$ in $\mathbb{Z}_3$ such that the quotient ring,

\begin{align*} R:= \frac{\mathbb{Z}_3\lbrack x \rbrack}{(x^3+x^2+bx+1)}, \end{align*}is a field.

I would appreciate a double check on my thinking.

Since $\mathbb{Z}_3$ is commutative, $R$ will be a field if and only if the ideal $(x^3+x^2+bx+1)$ is maximal. Since $\mathbb{Z}_3$ is a field, $\mathbb{Z}_3\lbrack x\rbrack$ is a PID, hence any irreducible element in $\mathbb{Z}_3\lbrack x\rbrack$ will generate a maximal ideal. Thus, it suffices to find all $b$ such that $x^3 + x^2 + bx + 1$ is irreducible. Since this polynomial is degree 3, it is sufficient to show that it has no roots in $\mathbb{Z}_3$. Evaluating the polynomial for $x = 0$ will never have a root. For the $x = 1$ and $x = 2$, we see: \begin{align*} b+3 &\equiv 0 \textrm{ mod } 3 \implies b \equiv -3 \textrm{ mod } 3 \equiv 0 \textrm{ mod } 3\\ b + 13 &\equiv 0 \textrm{ mod } 3 \implies b \equiv -13 \textrm{ mod } 3 \equiv 1\textrm{ mod } 3. \ \end{align*}

Since $(x^3 + x^2 + bx + 1)$ is reducible if $b \equiv 0 \textrm{ mod } 3$ or $1 \textrm{ mod }$, the only value making $R$ a field is if $b = 2$.

$\endgroup$
4
  • 1
    $\begingroup$ Same comment as above, plus 2 remarks: 1) since you agreed with the correction, you should edit your post to register it, for future readers 2) when asking for a solution verification, you are supposed to specify which step you are doubting and why: math.stackexchange.com/tags/solution-verification/info "For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context)..." $\endgroup$ Commented Jan 3 at 7:37
  • 1
    $\begingroup$ These are answers to the question, not just a comment. Per site policy, comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore I suggest that you post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. Also notice that comments are not indexed by the full text search, don't have any revision history, can only be edited for 5 minutes and allow only limited markup. $\endgroup$ Commented Jan 3 at 9:54
  • 1
    $\begingroup$ @AnneBauval Editing corrections from answers into a 'solution-verification'-question changes the question fundamentally, invalidating the answers. This is against site policy. $\endgroup$
    – Servaes
    Commented Jan 3 at 10:00
  • $\begingroup$ 1) I did not suggest to edit a correction from the (point 2 of the) answer, but from the (1 hour before) first comment, agreed by the OP (these 2 first comments disappeared, and so did the gratefulOP). And that was a minor correction. (So would point 1, which is only stylistic). 2) I noticed that very short answers to solution-verification questions (saying essentially: "yes, correct") are generally frowned upon on this site. 3) I won't anyway re-type my (not by me) deleted comments, because the OP didn't follow the rules about this type of question, recalled above in my (now) "first" comment. $\endgroup$ Commented Jan 3 at 15:22

1 Answer 1

1
$\begingroup$

Your approach is good, just two details are off:

  1. You state that any irreducible element will generate a maximal ideal. But is every maximal ideal generated by an irreducible element?
  2. Plugging in $x=2$ should yield the congruence $2b+13\equiv0\pmod3$.
$\endgroup$
3
  • $\begingroup$ 1) could be just a comment. 2) was mentionned before, in the first comment. $\endgroup$ Commented Jan 3 at 7:51
  • 3
    $\begingroup$ @AnneBauval I don't understand. The question asks for feedback on the attempted solution. That is exactly what I give here. $\endgroup$
    – Servaes
    Commented Jan 3 at 9:42
  • 4
    $\begingroup$ @AnneBauval Comments should not be used for giving answers. See How do comments work for more information. It is good that someone wrote an answer. $\endgroup$ Commented Jan 3 at 9:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .