I know that $1$ is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.

However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=0$ is a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?

(I know that this would violate the Fundamental Theorem of Arithmetic.)

Edit: Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see wikipedia).

This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see wikipedia).

So, my question is now: Why do we make this awkward convention?

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    Every number divides zero. – Asaf Karagila Aug 31 '10 at 09:04
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    @Asaf Karagila: That's true and it's the reason why 0 being prime would violate the Fundamental Theorem of Arithmetic. It's just that I always thought that one should think of prime numbers as prime ideals in Z. – Rasmus Aug 31 '10 at 09:17
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    "Prime numbers" should be called "maximal numbers". – Pierre-Yves Gaillard Aug 31 '10 at 10:32
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    What I meant by my previous comment is this. The notions of "domain" and "field", and thus the notions of "prime" and "maximal" ideal are, by far, important enough to deserve a name. And I see no reason to change the name they have. The obvious conclusion is that a ring element should be called "prime" (resp. "maximal") if so is the ideal it generates. Of course I know that it is far too late for this "obvious conclusion" to have any chance to prevail. – Pierre-Yves Gaillard Aug 31 '10 at 14:24

3 Answers3


You have a point here: absolutely we want to count $(0)$ as a prime ideal in $\mathbb{Z}$ -- because $\mathbb{Z}$ is an integral domain -- whereas we do not want to count $(1)$ as being a prime ideal -- because the zero ring is not an integral domain (which, to me, is much more a true fact than a convention: e.g., every integral domain has a field of fractions, and the zero ring does not).

I think we do not want to call $0$ a prime element because, in practice, we never want to include $0$ in divisibility arguments. Another way to say this is that we generally want to study factorization in integral domains, but once we have specified that a commutative ring $R$ is a domain, we know all there is to know about factoring $0$: $0 = x_1 \cdots x_n$ iff at least one $x_i = 0$.

Here is one way to make this "ignoring $0$" convention look more natural: the notions of factorization, prime element, irreducible element, and so forth in an integral domain $R$ depend entirely on the multiplicative structure of $R$. Thus we can think of factorization questions as taking place in the cancellative monoid $(R \setminus 0,\cdot)$. (Cancellative means: if $x \cdot y = x \cdot z$, then $y = z$.) In this context it is natural to exclude zero, because otherwise the monoid would not be cancellative. Contemporary algebraists often think about factorization as a property of monoids rather than integral domains per se. For a little more information about this, see e.g. Section 4.1 of http://math.uga.edu/~pete/factorization2010.pdf.

Pete L. Clark
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  • When you study Dedekind rings, you prove that any nonzero ideal is, in a "unique" way, a product of maximal ideals, but you still call (0) a prime ideal. – Pierre-Yves Gaillard Aug 31 '10 at 15:11
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    You support your claim that the non-integraldomainness of the zero ring is not a convention in that the zero ring doesn't have a field of fractions, but that is a convention, too, no? The zero field is not a field because we insist that $1\neq0$ in a field. – Mariano Suárez-Álvarez Aug 31 '10 at 15:12
  • I agree with Mariano. I feel that insisting on having 0 not equal to 1 for domains and fields, but not for rings, is the right convention, but I'm unable to say why... – Pierre-Yves Gaillard Aug 31 '10 at 15:23
  • To answer (at least partially) my own question: It's natural to call "maximal" an ideal which is maximal AMONG THE PROPER IDEALS (otherwise the notion would have no interest), and thus to call "field" the quotient of a ring by a maximal ideal. Now, as Pete says, if you want each domain to have a fraction field, you don't want a zero domain. – Pierre-Yves Gaillard Aug 31 '10 at 16:18
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    @MSA, PYG: ultimately a convention is just a certain kind of definition, and we obviously can't do away with definitions. The best we can do is try to choose definitions which handle degenerate cases gracefully (the French are especially good at that, I find). In this case, the setup that I find most elegant at the moment is that the spectrum of the zero ring is the empty space, which is not irreducible because it has zero irreducible components (not one). A field, of course, has a one point spectrum. No points versus one point -- that's like night versus day, right? :) – Pete L. Clark Sep 01 '10 at 09:04
  • Dear Pete: Wonderful comment!!! – Pierre-Yves Gaillard Sep 01 '10 at 17:00

There are good reasons behind the convention of including $(0)$ as a prime ideal but excluding $(1).\ $ First, we include zero as a prime ideal because it facilitates many useful reductions. For example, in many ring theoretic problems involving an ideal $\, I\,$, one can reduce to the case $\,I = P\,$ prime, then reduce to $\,R/P,\,$ thus reducing to the case when the ring is a domain. In this case one simply says that we can factor out by the prime $ P\,$, so w.l.o.g. assume $\, P = 0\,$ is prime, so $\,R\,$ is a domain. For example, I've appended to the end of this post an excerpt from Kaplansky's classic textbook Commutative Rings, section $1\!\!-\!\!3\!:\,G$-Ideals, Hilbert Rings, and the Nullstellensatz.

Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $(1)$ or, equivalently, why don't we permit the zero ring as a domain? There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\,1 = 0.\ $ More importantly, it implies that the unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\, R = 0)\,$ to the many arguments involving units and unit groups. For a more general perspective it's worth emphasizing that the usual rules for equational logic are not complete for empty structures so that is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this thread for details).

Below is the promised Kaplansky excerpt on reduction to domains by factoring out prime ideals. I've explicitly emphasized the reductions e.g. reduce to....

Let $\, I\,$ be any ideal in a ring $\, R.\,$ We write $\, R^{*}\,$ for the quotient ring $\, R/I.\,$ In the polynomial ring $\, R[x]\,$ there is a smallest extension $\, IR[x]\,$ of $\, I.\,$ The quotient ring $\, R[x]/IR[x]\,$ is in a natural way isomorphic to $\, R^*[x].\,$ In treating many problems, we can in this way reduce to the case $\, I = 0,\,$ and we shall often do so.

THEOREM $28$. $\,$ Let $\, M\,$ be a maximal ideal in $\, R[x]\,$ and suppose that the contraction $\, M \cap R = N\,$ is maximal in $\, R.\ $ Then $\, M\,$ can be generated by $\, N\,$ and one more element $\, f.\ $ We can select $\, f\,$ to be a monic polynomial which maps $\!\bmod N\,$ into an irreducible polynomial over the field $\, R/N.\ $

Proof. $\,$ We can reduce to the case $\, N = 0,\,$ i. e., $\, R\,$ a field, and then the statement is immediate.

THEOREM $31$. $\,$ A commutative ring $\, R\,\,$ is a Hilbert ring if and only if the polynomial ring $\, R[x] \,\,$ is a Hilbert ring.

Proof. $\,$ If $\, R[x]\,$ is a Hilbert ring, so is its homomorphic image $\, R\,$. Conversely, assume that $\, R\,$ is a Hilbert ring. Take a G-ideal $\, Q\,$ in $\, R[x]\,$; we must prove that $\, Q\,$ is maximal. Let $\, P = Q \cap R\,$; we can reduce the problem to the case $\, P = 0,\,$ which, incidentally, makes $\, R\,$ a domain. Let $\, u\,$ be the image of $\, x\,$ in the natural homomorphism $\, R[x] \to R[x]/Q.\,$ Then $\, R[u]\,$ is a G-domain. By Theorem $23$, $\,u\,$ is algebraic over $\,R\,$ and $\,R\,$ is a G-domain. Since $\,R\,$ is both a G-domain and a Hilbert ring, $\,R\,$ is a field. But this makes $\, R[u] = R[x]/Q\,$ a field, proving $\, Q\,$ to be maximal.

Bill Dubuque
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  • Thank you for your detailed answer. – Rasmus Aug 31 '10 at 14:18
  • Dear Bill, could you please explain in more detail why we can reduce to the cases $N=0$ and $P=0$? If you have more instructive examples of modding out by primes, I would love to see. – Arrow May 05 '20 at 17:10
  • @Arrow I have linked many further examples here (some simpler than the above) - see the "Linked" questions in the sidebar. I recommend that you peruse them for further motivation. – Bill Dubuque May 05 '20 at 17:25
  • @Arrow [Another example](https://math.stackexchange.com/a/210942/242) (prime ideals are maximal iif every element $r$ satisfies $\,r^n = r\,$ for some $\,n>1)$ – Bill Dubuque May 17 '20 at 17:30
  • @Arrow Another example is [here](https://math.stackexchange.com/a/173283/242) (prime ideals of a polynomial ring). – Bill Dubuque May 21 '20 at 23:21

Generally we make nice conventions because they make the statements of theorems nice. The theorem relevant to prime ideals is that $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain. The theorem relevant to prime elements is prime factorization (when it holds).

These two concepts almost coincide for principal ideals, but we must distinguish between the generic point $(0)$ and closed points, and there are good reasons for doing this. (The zero ideal, for example, can't occur in the factorization of a nonzero ideal in a Dedekind domain.)

Qiaochu Yuan
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