# Prime ideals and irreducible ideals

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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Prime ideals should also satisfy $I\neq R$. – Arturo Magidin Jan 20 '12 at 19:48
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. – user23214 Jan 20 '12 at 19:53
@user23214 See, this is the problem with trying to learn Algebraic Geometry with a less-than-desired knowledge of commutative algebra. Thanks. – Matt Jan 20 '12 at 19:58
@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. – Arturo Magidin Jan 20 '12 at 20:09
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. – Thomas Andrews Jan 20 '12 at 20:29

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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