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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?

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  • $\begingroup$ 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. $\endgroup$ Jul 12, 2014 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, 2014 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, 2014 at 18:03
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    $\begingroup$ 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. $\endgroup$ Jul 12, 2014 at 18:08
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    $\begingroup$ @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. $\endgroup$ Jul 12, 2014 at 18:46

3 Answers 3

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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.

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    $\begingroup$ In fact you prove that every PID is a UFD exactly in this way. $\endgroup$
    – Crostul
    Jul 12, 2014 at 18:11
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    $\begingroup$ 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, 2014 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, 2014 at 18:14
  • $\begingroup$ @Andrew D: I totally agree with you, I was stressing just on what you said. $\endgroup$
    – Crostul
    Jul 12, 2014 at 18:17
  • $\begingroup$ @Crostul Ah, I see. $\endgroup$
    – Andrew D
    Jul 12, 2014 at 18:18
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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.

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    $\begingroup$ @VHP I recomomend that you give some consideration to this viewpoint, since it may prove conceptually useful later in your studies. $\endgroup$ Jul 12, 2014 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, 2014 at 18:37
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    $\begingroup$ @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. $\endgroup$ Jul 12, 2014 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, 2017 at 3:23
  • $\begingroup$ @Maggie In a PID every ideal is principal. $\endgroup$ Apr 7, 2017 at 3:40
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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 $\mathbb{Q}[x]/P\cong\mathbb{Q}$, 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 $I=R$.

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  • $\begingroup$ 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. $\endgroup$ Aug 2, 2015 at 0:12
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    $\begingroup$ @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, 2015 at 7:58
  • $\begingroup$ Oh, I see. Thank you very much :) $\endgroup$ Aug 3, 2015 at 18:41
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    $\begingroup$ @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, 2017 at 23:47
  • $\begingroup$ @Cici Since $I$ contains an invertible element, namely $a$, we conclude $I=R$. $\endgroup$
    – egreg
    Sep 17, 2017 at 8:28

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