12
$\begingroup$

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal

Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which every ideal $A$ is of the form $\langle a \rangle = \{ar~~ | ~~r \in R\}$

Let $ \langle a \rangle $ be a prime ideal $\implies R/A $ is an integral domain.

A finite integral domain is a field . Hence, if we prove that $R/A$ is finite, then $R/A$ is a field $\implies A$ is a maximal ideal.

Now, $\langle a \rangle = \{ar~~ | ~~r \in R\}$

Since, R is an integral domain, there are no zero divisors and cancellation is allowed $\implies ar_1 = ar_2 \implies r_1=r_2 \implies ar_i$ maps to a different member of $R$ for each different $r_i \implies \langle a \rangle$ represents the elements of $R$ in some random order.

$\implies \langle a \rangle = R$ and hence, $R/A \approx {0}$ is finite and hence $A$ is maximal.

Is my attempt correct?

|cite|improve this question
$\endgroup$
  • $\begingroup$ No, this is impossible. Consider $\Bbb Q[x]$ which is a PID because it is a Euclidean domain (polynomials over a field). Irreducible polynomials generate the maximal ideals, but the quotient is not a finite field. $\endgroup$ – Adam Hughes Jul 12 '14 at 18:01
  • $\begingroup$ So you proved $\;\langle a\rangle=R\;$ and thus any prime ideal is the whole ring? This is obviously wrong...And besides this, $\;A=\langle a\rangle =R\implies R/A=R/R\cong\{0\}\;$ $\endgroup$ – Timbuc Jul 12 '14 at 18:01
  • $\begingroup$ @AdamHughes and Timbuc can you please pin point in the proof where a mistake could have been committed? $\endgroup$ – MathMan Jul 12 '14 at 18:03
  • 1
    $\begingroup$ Your error is assuming that just because the map: $$\begin{cases}f_a: R\to R \\ r\mapsto ra\end{cases}$$ is bijective just because it is injective (domain implies injective). In fact, injective implies bijective if $R$ is finite, but not all integral domains are finite, take $\Bbb Z$, look at the multiplication by $2$ map, you get all the even numbers, but it clearly does not permute the elements bijectively. $\endgroup$ – Adam Hughes Jul 12 '14 at 18:08
  • 1
    $\begingroup$ @VHP glad to help. I also read about how you hadn't seen UFDs, so I made an edit which should help you out for where you are in your book. $\endgroup$ – Adam Hughes Jul 12 '14 at 18:46
16
$\begingroup$

To prove this we use that $PID$ are UFDs. Then Let $\mathfrak{p}=(p)$ be a prime ideal of $R$. Assume there is an ideal $\mathfrak{p}\subseteq\mathfrak{m}=(m)\subseteq R$. Then we would have that $m|p$, but then by the defintion of a prime element of a UFD this means that $m=p$ or $m=u$, a unit. Hence $\mathfrak{m}=(p)$ or $R$, proving maximality.

Edit: Since the op hasn't seen UFDs yet, here's a quick way around that:

Becaues $(p)\subseteq (m)$ we have that there is a $b\in R$ so that $ap=bm$ for every $ap\in (p)$. In particular $p=b_0m$. But then as $p$ is prime, either $m$ or $b_0$ must be a unit. the first case implies $\mathfrak{m}=R$ the second implies $\mathfrak{m}=\mathfrak{p}$, hence $\mathfrak{p}$ is maximal.

|cite|improve this answer
$\endgroup$
  • 4
    $\begingroup$ In fact you prove that every PID is a UFD exactly in this way. $\endgroup$ – Crostul Jul 12 '14 at 18:11
  • 2
    $\begingroup$ Isn't this often used in the proof that PID's are UFD's? I mean, you can very easily adapt the proof so you aren't using anything UFD specific (prove the chain that if $r$ is non-zero/a non-unit, $r$ irreducible $\implies$ $(r)$ prime $\implies$ $(r)$ maximal $\implies$ $r$ irreducible) $\endgroup$ – Andrew D Jul 12 '14 at 18:12
  • $\begingroup$ @Crostul You'd also need to prove that PID's are FD's, and also prove that the property that the above property and being a FD is equilivant to it being a UFD (where we usually define it as saying that prime factorisations exist and are unique up to units) $\endgroup$ – Andrew D Jul 12 '14 at 18:14
  • $\begingroup$ @Andrew D: I totally agree with you, I was stressing just on what you said. $\endgroup$ – Crostul Jul 12 '14 at 18:17
  • $\begingroup$ @Crostul Ah, I see. $\endgroup$ – Andrew D Jul 12 '14 at 18:18
13
$\begingroup$

Hint $ $ Notice that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supset (b)\iff a\mid b.\,$ Therefore

$\qquad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\!\! (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ &\iff& p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ &\iff& p\ \ \text{ is irreducible}\\ &\iff& \!\!(p)\ \text{ is prime, } \ \text{by PID} \Rightarrow\text{UFD,}\ \text{ so ireducible = prime } \end{eqnarray}$

Remark $\ $ PIDs are the UFDs of dimension $\le 1,\,$ i.e. where all prime ideals $\ne 0\,$ are maximal.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ @VHP I recomomend that you give some consideration to this viewpoint, since it may prove conceptually useful later in your studies. $\endgroup$ – Bill Dubuque Jul 12 '14 at 18:29
  • $\begingroup$ Surely, I will. I just completed the Factor rings chapter in Gallian. It hasn't yet introduced the Unique Factorisation Domain as of yet, but will try to relate what you said when I start reading about it. $\endgroup$ – MathMan Jul 12 '14 at 18:37
  • 1
    $\begingroup$ @VHP You need only the direction $(\Leftarrow)$ which uses only the simple direction of the final equivalence, i.e. that primes are irreducible. So no knowledge of UFDs is needed for that. $\endgroup$ – Bill Dubuque Jul 12 '14 at 19:17
  • $\begingroup$ Could you explain the first equivalence that $(p)$ is maximal is equivalent to $(p)$ has no proper container $(a)$? Why isn't there some other type of ideal (not principal) that contains $(p)$? $\endgroup$ – Maggie Mak Apr 7 '17 at 3:23
  • $\begingroup$ @Maggie In a PID every ideal is principal. $\endgroup$ – Bill Dubuque Apr 7 '17 at 3:40
11
$\begingroup$

Your approach can't work. The ring $\mathbb{Q}[x]$, where $\mathbb{Q}$ is the field of rational numbers, is a PID. However, for the prime ideal $P=\langle x\rangle$ the quotient is $\mathbb{Q}[x]/I\cong\mathbb{Q}$, which is infinite.

Suppose $P$ is a nonzero prime ideal and $P=\langle p\rangle$. Now, let $I=\langle a\rangle$ be an ideal such that $\langle p\rangle\subseteq\langle a\rangle$. In particular $$ p=ab $$ so $ab\in P$. Therefore, by definition of prime ideal, either $a\in P$ or $b\in P$.

If $a\in P$, then $\langle a\rangle\subseteq\langle p\rangle$, whence $I=P$.

If $b\in P$, then $b=pc$, so $p=ab=apc$ from which $1=ac$ and $a$ is invertible, whence $I=R$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ How do we know that $p=apc$ imples that $1=ac$? I mean, we don't know for sure that $p$ has a multiplicative inverse so we can not simply cancel it out. So why does it still allowed to cancel it? Any clarification? thank you in advance. $\endgroup$ – Fawzy Hegab Aug 2 '15 at 0:12
  • 3
    $\begingroup$ @MathsLover We are in an integral domain: $p=apc$ is the same as $p(1-ac)=0$ and, since $p\ne0$, we conclude $1-ac=0$. $\endgroup$ – egreg Aug 2 '15 at 7:58
  • $\begingroup$ Oh, I see. Thank you very much :) $\endgroup$ – Fawzy Hegab Aug 3 '15 at 18:41
  • $\begingroup$ @egreg I'm a bit stuck on the last line of your proof, with $b \in P$.... I understand that $c = a^{-1}$, but why does that mean $I = R$? Would really appreciate the clarification! $\endgroup$ – Cici Sep 16 '17 at 23:47
  • $\begingroup$ @Cici Since $I$ contains an invertible element, namely $a$, we conclude $I=R$. $\endgroup$ – egreg Sep 17 '17 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.