# Why are prime ideals proper?

As children we all learn this erroneous definition of a prime number: “a number $$n\in \Bbb N$$ is prime iff it’s only divided by one and itself”. Well that’s fine until the teacher asked us for examples. Some common responses might have been “2, 7, 3, 4, 1-“ actually no—one isn’t prime. When asked why, the teacher may respond with something along the lines of, “well, the actual definition of prime number is for some ring $$R$$, an element $$x\in R$$ is prime iff $$\langle x\rangle\subseteq R$$ is a prime ideal.”*

Well, I’m in college now and know that and ideal $$I\subseteq R$$ of a ring $$R$$ is prime iff for any $$a,b\in R$$ if $$ab\in I$$ then $$a\in I$$ or $$b\in I$$ and $$I$$ is proper. Well this makes sense and all so now let’s see why one isn’t prime. It’s because prime ideals must be proper and $$\langle 1\rangle =R$$ always. Well that’s disappointing. Why must prime ideals be proper?

I’d want the historical reason behind this definition (if available) and some other conveniences from this exclusion and if you want to be contrarian, some inconveniences.

Here are some guesses:

Much like $$1$$ has an empty factorization into primes and so having one be prime invalidates the unique composition into primes of positive integers, $$\langle 1\rangle$$ can also be gotten as an empty product as it’s the identity in the semi ring of ideals with the usual operations.

If prime ideals are maximal, then having $$\langle 1\rangle$$ is a bit of an outlier

Something to do with algebraic geometry

*OK maybe that never happened but the actual explanation “the prime needs two divisors, having this repeat doesn’t work” wasn’t very convincing to me.

• It might be (at least partially) explained by the observation that $\mathfrak{p}$ is prime if and only if $R / \mathfrak{p}$ is a non-zero integral domain. It is convenient to define that $\{0\}$ is not an integral domain, and therefore equivalently that $R$ is not prime. Commented Jul 17 at 0:40
• Some answers from "Why $1$ is not a prime number?" also apply here, e.g. here. Commented Jul 17 at 1:00
• @J.S. I’d considered the integral domain answer but it has the seemingly arbitrary exclusion of the zero ideal. The convenience of excluding the zero ideal seems a bit too far off from prime ideals to be a satisfactory answer to me. Commented Jul 17 at 1:02
• Do you mean the zero ring? I wouldn't say it's arbitrary; my view of integral domains and prime ideals is that they are two sides of the same coin. In any case, you might at least find more reasons indirectly by searching for "why $\{0\}$ is not an integral domain". I thought of an explicit example as well; if $R$ is a prime ideal, then $S = R \setminus R$ is not a multiplicative set, which is another inconvenience if you want to localise at prime ideals. Commented Jul 17 at 1:09
• It's not "far off" since they are equivalent statements. Reasons why $\{0\}$ is not a domain are already explained in the link I gave above (and elsewhere here). In any case it is a matter of convention. One normally chooses definitions that prove most convenient in any given context. Commented Jul 17 at 1:12

This is a central example of a phenomenon the nLab calls too simple to be simple. Some other related examples: the empty graph and empty topological space are not connected, the zero module over a ring is not simple. The nLab has other examples. Typically examples here are confusing because people give definitions that don't correctly specialize to "degenerate" cases.

Sticking to just the case of prime numbers to start with, here is one way to define a prime number that transparently excludes $$1$$: associated to any positive integer $$n$$ we can consider its poset of divisors $$\{ d : d \mid n \}$$. If $$n$$ has unique factorization $$\prod p_i^{n_i}$$ then this poset is correspondingly a product $$\prod \{ 0 < \dots < n_i \}$$ of finite chains, one chain for each prime divisor of $$n$$.

Definition: A positive integer $$n$$ is prime iff its poset of divisors is a chain $$\{ 1 \mid n \}$$ of length exactly $$2$$.

So $$1$$ isn't prime because its poset of divisors is a chain of length $$1$$. Chains of length $$1$$ and chains of length $$2$$ behave quite differently!

Now let's discuss prime ideals. Of course there's a historical motivation here: we want ideals in a Dedekind domain, such as the ring of integers $$\mathcal{O}_K$$ of a number field $$K$$, to have a unique factorization into prime ideals, and this requires that the unit ideal not be considered prime. I would guess that this is the historical motivation, since this was the historical motivation behind the definition of ideals in the first place (Kummer called them "ideal numbers").

But from a modern perspective this is a pretty specific class of rings; also, since nonzero prime ideals of a Dedekind domain are maximal, thinking only about Dedekind domains doesn't really distinguish between prime and maximal ideals, either. We ought to be able to say something more general about prime ideals in arbitrary commutative rings. My preferred definition of a prime ideal is that an ideal $$P$$ of a commutative ring $$R$$ is prime iff $$R/P$$ is an integral domain. So the question of why prime ideals have to be proper is equivalent to:

Why must we exclude the zero ring $$0 = \{ 0 \}$$ from the definition of an integral domain?

There are several different answers you could give to this, off the top of my head.

1. One way to define integral domains is that they are exactly the subrings of fields. So this further reduces the question to: why must we exclude the zero ring from the definition of a field? There are again several ways to respond to this, but for example, one way to define a field is that a field $$F$$ is a commutative ring such that $$F \setminus \{ 0 \}$$ is a group under multiplication. This naturally forces $$1 \neq 0$$.

2. It's also worth discussing maximal ideals here, since one way to define maximal ideals is that an ideal $$m$$ is maximal iff the quotient $$R/m$$ is a field. Here is a different way to define fields which directly generalizes the definition I gave above of prime numbers: a commutative ring $$R$$ is a field iff its poset of ideals is a chain $$\{ 0 \subsetneq 1 \}$$ of length exactly $$2$$. So this is also a way to explain why maximal ideals must be proper.

3. A commutative ring $$R$$ being an integral domain roughly corresponds to its spectrum $$\text{Spec } R$$ being irreducible (strictly speaking we need $$R$$ to be reduced for this to be an if-and-only-if). Irreducibility is a strong connectedness condition, and so disallowing the zero ring from being an integral domain (equivalently, forcing the spectrum of the zero ring to be empty) is closely related to the idea that the empty topological space is not connected. In turn, one way to justify the disconnectedness of the empty topological space is that a connected topological space should be a topological space that has exactly $$1$$ connected component, while the empty space has $$0$$ connected components.

4. Some definitions in algebra make more sense when you try to phrase them in terms of $$n$$-ary operations rather than just binary operations. An $$n$$-ary way to describe an integral domain is that a commutative ring $$R$$ is an integral domain if $$R \setminus \{ 0 \}$$ is closed under $$n$$-ary multiplication for all $$n$$. Setting $$n = 0$$ gives that $$R \setminus \{ 0 \}$$ must be closed under empty products, and the empty product is $$1$$. So this forces $$1 \neq 0$$. We see from this line of reasoning that the usual definition of an integral domain, which considers $$n = 2$$ only, imposes this condition for all values of $$n$$ except $$0$$, but we get a more uniform description that correctly excludes the zero ring by writing down a condition for all $$n$$ simultaneously.

This last explanation is new to me but I think I like it the best. It's closely related to the idea that empty products explain why rings should have multiplicative identities; this is not an extra axiom but a special case of $$n$$-ary associativity, as explained e.g. by Bjorn Poonen in Why all rings should have a $$1$$.

• Among other things, my conviction is that this is not a matter of convention, and that with the correct definitions we should not have to exclude $1$, the empty space, etc. as special cases. Commented Jul 17 at 2:18
• I agree with the finite over n=0,2 approach like with finite unions for topological spaces instead of the empty set and binary unions. Think I also like that explanation the best. Commented Jul 17 at 3:39
• @LaveCave: I think Bjorn Poonen also proposes a direct explanation of why prime ideals should be proper in his paper (see section 4). If $A$ is a ring and $I\subseteq A$ is an ideal, then it seems to natural to say that $I$ is prime if the complement of $I$ in $A$ is closed under finite products. Thus, the improper ideal is not prime, since the empty set is not closed under finite products (recall that $1$ is the empty product).
– Joe
Commented Jul 17 at 12:57
• I’d say that a more direct way to do this would be to define a prime ideal as an ideal $I\subseteq R$ such that for any finite index set $J$, $\prod_{j\in J} a_j\in I\to \bigvee_{j\in J} a_j\in I$. In the n=0 case, it gives that $1\in I\to \bot$ because the empty disjunction is false. This is just the statement that $I\neq R$. (I didn’t see Joe’s comment at the time of writing. This was a response to my previous comment). Commented Jul 17 at 14:16