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I have the following definitions:


An ideal $I$ is prime if, whenever $ab \in I$, either $a \in I$ or $b \in I$.

An ideal $J$ is irreducible if, whenever $J = I_1 \cap I_2$ for ideals $I_1 $ and $ I_2$, $J \subseteq I_1$ or $J \subseteq I_2$. EDIT: As per Arturo Magidin's correction, we must have $ J = I_1$ or $J = I_2$.


Under what conditions are prime ideals irreducible ideals? What about the converse?

I'd guess that:

i) prime ideals are always irreducible (mirroring the result for elements of a ring) [provided we're in an integral domain]

ii) irreducible ideals aren't always prime. Are there any nice examples?

iii) In certain types of ring, irreducible ideals are always prime (e.g. a UFD or stronger). I'd guess this mirrors the result for elements

iv) The ideal $ \langle a\rangle$ is prime iff $a$ is prime, and similarly in the irreducible case. So in PIDs, prime ideals are irreducible ideals and they are generated by a (prime) irreducible.

Am I correct? How would I go about proving these (especially iii)?

Thank you.

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    $\begingroup$ Prime ideals should also satisfy $I\neq R$. $\endgroup$ – Arturo Magidin Jan 20 '12 at 19:48
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    $\begingroup$ In Noetherian rings, it can be shown that every irreducible ideal is primary. (for instance, Atiyah-MacDonald thm. 7.12) So in general, irreducible ideals shouldn't even always be primary. $\endgroup$ – user23214 Jan 20 '12 at 19:53
  • $\begingroup$ @user23214 See, this is the problem with trying to learn Algebraic Geometry with a less-than-desired knowledge of commutative algebra. Thanks. $\endgroup$ – Matt Jan 20 '12 at 19:58
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    $\begingroup$ @Matt: Your definition of "irreducible" is incorrect: the correct definition is "$J=I_1\cap I_2$ implies $J=I_1$ or $J=I_2$"; equality in the conclusion, not merely inclusion. Otherwise, every principal ideal would be irreducible. $\endgroup$ – Arturo Magidin Jan 20 '12 at 20:09
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    $\begingroup$ If $J=I_1\cap I_2$ is prime, and $a_i\in I_i\setminus J$ for $i=1,2$, then $a_1a_2\in I_1\cap I_2 = J$, so either $a_1\in J\subset I_1$ or $a_2\in J\subset I_2$, contradicting our assumption. $\endgroup$ – Thomas Andrews Jan 20 '12 at 20:29
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  1. Assume that $J\neq R$ is prime, and $J=I_1\cap I_2$. Then $I_1I_2\subseteq I_1\cap I_2 = J$; since $J$ is prime, then $I_1\subseteq J$ or $I_2\subseteq J$. But then, if we have $I_1\subseteq J = I_1\cap I_2$, so $I_1\subseteq I_2$; in particular, $J=I_1\cap I_2 = I_1$, so we conclude that $J=I_1$; analogously, if $I_2\subseteq J$, then we conclude that $J=I_2$, so the condition for irreducibility is satisfied. You don't need $R$ to be an integral domain. (I didn't even need $R$ to be commutative, since the proof above is ideal-wise, not element-wise!)

  2. Take $R=\mathbb{Z}$, $J=(p^n)$ where $p$ is a prime and $n\gt 0$. If $(p^n)=(a)\cap (b) = (\mathrm{lcm}(a,b))$ then we must have $a=\pm p^i$, $b=\pm p^j$ for some $i,j$, $0\leq i,j\leq n$; and we must have $\max\{i,j\}=n$. Hence $(p^n)=(a)$ or $(p^n)=(b)$, so $(p^n)$ is irreducible. However, it is only prime if $n=1$.

  3. Is incorrect, as witnessed by the example in 2; even Euclidean is not sufficient for irreducible to imply prime.

    But in a PID (or more generally, a Noetherian ring), irreducible and radical does imply prime. For in a PID (or as noted by user23214, in a Noetherian ring), irreducible implies primary; if $J$ is primary and radical, then it is prime. But this is probably too much of a sledgehammer condition.

    (An ideal $I$ is primary if $I\neq R$ and $xy\in I$ implies $x\in I$ or $y^n\in I$ for some $n\gt 0$; an ideal $I$ is radical if $I=\sqrt{I}$, if $x^n\in I$ for some $n\gt 0$ implies $x\in I$.)

  4. Same problem for the statement about irreducible elements/ideals. But in a PID, an ideal generated by an irreducible is necessarily irreducible (if $(m)=(a)\cap(b)$, then $a|m$, and irreducibility of $m$ implies $(a)=(1)$ or $(a)=(m)$; similarly with $b$, and we cannot have both $(a)=(1)$ and $(b)=(1)$), though the converse need not hold, as witnessed by the example in 2.

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In any commutative ring with unity, prime ideals are proper (by definition), irreducible, and radical. As noted in comments and answers above, the converse is true: proper, irreducible, and radical ideals are prime. However, this can be done without a "sledgehammer"; it does not require Noetherianness, primary ideals, etc. More precisely:

Claim: Let $R$ be a commutative ring with unity. Let $I \subsetneq R$ be a proper ideal. Assume that $I$ is irreducible (if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$) and $I$ is radical (if $x^n \in I$ for any $n \geq 1$, then $x \in I$). Then $I$ is prime ($I \neq (1)$, and if $ab \in I$, then $a \in I$ or $b \in I$).

Remark: Note that the unit ideal $I=(1)=R$ is irreducible and radical, but not prime. In $R=\mathbb{Z}$, $I=(4)$ is proper and irreducible, but not prime; $I=(6)$ is proper and radical, but not prime. So, no two out of $\{$proper, irreducible, radical$\}$ are sufficient to get primeness—which is no surprise, since primeness implies proper, irreducible, and radical (all three).

Anyway, the hypotheses are necessary, let us proceed to the proof that they are sufficient.

Proof of Claim: By contrapositive. Let $I$ be a proper ideal, that is $I \neq (1)$. Assume $I$ is radical but not prime; our goal is to show that $I$ is not irreducible. In general there are two ways that $I$ can fail to be prime: either $I=(1)$, or there are some elements outside $I$ whose product is in $I$. We are assuming that $I$ is a proper ideal, $I \neq (1)$, so it must be the second possibility.

So there are some elements $a,b \in R$ such that $ab \in I$ while $a,b \notin I$. Consider the ideals $I+(a)$ and $I+(b)$. Certainly $I \subseteq (I+(a)) \cap (I+(b))$. For the reverse inclusion, suppose $x \in (I+(a)) \cap (I+(b))$. Write $x = i_1 + r_1 a = i_2 + r_2 b$ for some $i_1,i_2 \in I$ and $r_1,r_2 \in R$. Then $$ x^2 = (i_1 + r_1 a)(i_2 + r_2 b) \in I $$ since, upon expansion, three of the terms involve $i_1$ or $i_2$ (or both) and the fourth term has $ab$, which is in $I$. Since $I$ is radical, $x \in I$. This proves $I = (I+(a)) \cap (I+(b))$. Since $a \notin I$, $I \neq I+(a)$; similarly $I \neq I+(b)$. Therefore, $I$ is not irreducible. $\square$

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  • $\begingroup$ Very nice! It would be clearer, I think, if you stated explicitly the assumption that $I$ is proper at the beginning of the poof. (The statement you're taking the contrapositive of is the one contained in the last two sentences of your claim, if my understanding is correct.) $\endgroup$ – Pierre-Yves Gaillard Apr 2 '18 at 22:50
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    $\begingroup$ @Pierre-YvesGaillard Thanks! I made some clarifications, starting from your suggestion. $\endgroup$ – Zach Teitler Apr 3 '18 at 4:14
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    $\begingroup$ Awesome! I'm sure you're aware of this, but let me point out to the reader a minor variant of the argument: The statement is easily reduced to the case $I=(0)$, which is slightly simpler. $\endgroup$ – Pierre-Yves Gaillard Apr 3 '18 at 12:37

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