$1.$ Find a maximal ideal and a prime ideal in $\mathbb Z_8[x]$

Attempt: Finding a maximal ideal, I am not sure how do I go about it. $\mathbb Z_8[x]$ is not a $PID$, so there's no use finding an irreducible polynomial as well.

EDIT: I am trying to show $2$ must be in all prime/maximal ideals

What should be the logical procedure in actually trying to find out a maximal ideal in any ring?

$2.$ Let $R=\{a/b~~|~~a,b \in \mathbb Z, 3 \nmid b\}$ . We need to find it's field of quotients.

Attempt: $R$ can be verified to be an integral domain, so, there exists a field $F$ called the field of quotients of $R$ which has a subring isomorphic to $D$

$R$ is not a field as $3/2$ is not invertible in $R$.

Let $p,q \in R$. Then let $p/q$ denote the equivalence class containing $(p,q)$

Then define $p/q + r/s = ps+rq/qs$ and $p/q \cdot r/s =pr/qs$

The set of these quotients form a field isomorphic to the set of rational whose denominators are not a multiple of $3$.

Am I correct?

Thank you for your help.

  • 2
    $\begingroup$ The ideal generated by $x$ is not prime, because the quotient ring is $\mathbb{Z}_8$ which is not a domain. Finding a maximal ideal in a general ring (commutative or not) requires the axiom of choice. $\endgroup$ – egreg Aug 3 '14 at 19:16
  • $\begingroup$ Ohh my bad. Removing it. $\endgroup$ – MathMan Aug 3 '14 at 19:19

Lots of questions here, so I'll focus on just one of them. One way of finding a maximal ideal in $\mathbb{Z}_8[x]$ is using the fact that an ideal $I$ is maximal if and only if $\mathbb{Z}_8[x]/I$ is a field.

First, recognize that $\mathbb{Z}_8/\langle 2 \rangle \cong \mathbb{Z}_2$, and so it follows that $\mathbb{Z}_8[x]/\langle 2 \rangle \cong \mathbb{Z}_2[x]$. What are the maximal ideals in $\mathbb{Z}_2[x]$? The corresponding maximal ideal in $\mathbb{Z}_8[x]$? Hint: The latter won't be principal.

I should add that all maximal ideals are prime ideals, so you've really answered two questions if you can answer this one.

Edit: If you wish to find an ideal that is prime, but not maximal, then apply the fact that an ideal $I$ in a ring $R$ is prime $\iff$ $R/I$ is an integral domain. Another hint for approaching this: $R/I$ cannot be finite since all finite integral domains are fields.

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    $\begingroup$ Ok, there we go. It's maximal because $\mathbb{Z}_2[x]/\langle x \rangle \cong \mathbb{Z}_2$ is a field. Now how can we get $\mathbb{Z}_2$ from $\mathbb{Z}_8[x]$? $\endgroup$ – Kaj Hansen Aug 3 '14 at 21:29
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    $\begingroup$ I think you mean $\phi:\mathbb{Z}_8[x] \rightarrow \mathbb{Z}_2$. I see where you're trying to go, but that won't work because your $\phi$ isn't a homomorphism. In particular, $\phi(f(x) + g(x)) = \phi(f(x)) + \phi(g(x))$ isn't always satisfied. If you can think of such a homomorphism, that's great, and you can do it that way! What I'm actually trying to get at here is that, if you have an ideal of a ring $R$ generated by, say $x$, and another ideal generated by $y$, then $(R/\langle x \rangle)/ \langle y \rangle \cong R/ \langle x, y \rangle$. $\endgroup$ – Kaj Hansen Aug 4 '14 at 4:00
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    $\begingroup$ Well, an ideal $I$ of a ring $R$ is maximal $\iff$ $R/I$ is a field. We're looking for an ideal to mod out $\mathbb{Z}_8[x]$ by so that we end up with a field, which we're taking to be $\mathbb{Z}_2$ in this case. $\endgroup$ – Kaj Hansen Aug 4 '14 at 6:24
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    $\begingroup$ Well that's one part of your question answered. Read the last paragraph of my original response. Now how can we manipulate $\mathbb{Z}_2[x]$ to convert it into a field? When you have an answer, think back to the last sentence of my second comment. $\endgroup$ – Kaj Hansen Aug 4 '14 at 6:47
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    $\begingroup$ Yeah, that's good. I'm slightly confused by notation, but the ideal generated by $2$ is prime in $\mathbb{Z}_8[x]$ as I think you're saying. To find a maximal ideal, apply the fact that $(R/\langle x \rangle)/\langle y \rangle \cong R/ \langle x, y \rangle$. You're very close. $\endgroup$ – Kaj Hansen Aug 4 '14 at 7:41

Hint for #1. Show that $2$ must be in all prime ideals. Also convince yourself of the fact that $\Bbb{Z}_8[x]/\langle 2\rangle\cong\Bbb{Z}_2[x]$. Can you find prime ideals/maximal ideals in the latter ring, and pull them back to the original ring?

Hint for #2. Can you think of a field that contains $R$? What is the smallest such field? Can you show that it is the field of quotients of $R$.

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    $\begingroup$ uhmm.. If $R$ is a commutative ring and $I$ is an ideal of $R$, then $R[x]/I[x] \approx (R/I)[x]$. Since $Z_8/ \langle 2 \rangle \approx Z_2 \implies Z_8[x]/ \langle 2 \rangle \approx Z_2[x]$ .So, the first part of the hint is done. Since $Z_2[x]$ is principal ideal domain, $\implies $ it suffices to find any irreducible polynomial in $Z_2[x]$. $x$ is such a polynomial? Hence $\langle x \rangle$ is a maximal ideal in $Z_2[x]$ . Am I on the correct path? for (2), the smallest field containing $R$ is $\mathbb Q$. But, I am confused due to the condition that the $3$ does not divide denominator? $\endgroup$ – MathMan Aug 3 '14 at 19:50
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    $\begingroup$ Looks good for part one (what's the preimage of that prime ideal in $\Bbb{Z}_8[x]$?). Can you write all the elements of $\Bbb{Q}$ as fractions with numerator and denominator both from $R$? $\endgroup$ – Jyrki Lahtonen Aug 3 '14 at 19:56
  • $\begingroup$ Sorry, I am stuck on finding the pre image of the ideal in $\mathbb Z_8[x]$ uhm, yeah, I think so. We can write all elements of $\mathbb Q$ using elements from $R$. $\endgroup$ – MathMan Aug 3 '14 at 20:11
  • $\begingroup$ Is the pre image $\langle 3x \rangle$? $\endgroup$ – MathMan Aug 3 '14 at 20:16
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    $\begingroup$ Well done, @VHP! $\endgroup$ – Jyrki Lahtonen Aug 4 '14 at 8:38

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