Is $\{0\}$ a field?

Question

Consider the set $F$ consisting of the single element $I$. Define addition and multiplication such that $I+I=I$ and $I \times I=I$ . This ring satisfies the field axioms:

• Closure under addition. If $x, y \in F$, then $x = y = I$, so $x + y = I + I = I \in F$.
• Closure under multiplication. $x \times y = I \times I = I \in F$
• Existence of additive identity. $\forall x \in F$ (i.e., for $x=I$), $x + I = x$, so $I$ is the additive identity.
• Existence of mulitiplicative identity. $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity.
• Additive inverse. $\forall x \in F, \exists y = I \in F: x + y = I$
• Multiplicative inverse. $\forall x \in F, \exists y = I \in F: x \times y = I$. However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth.
• Commutativity of addition. $\forall x, y \in F, x + y = I = y + x$
• Commutativity of multiplication. $\forall x, y \in F, x \times y = I = y \times x$
• Associativity of addition. $\forall x, y, z \in F$, $(x + y) + z = I + I = I$ and $x + (y + z) = I + I = I$, so $(x + y) + z = x + (y + z)$
• Associativity of multiplication. $\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$
• Distributivity of multiplication over addition. $I \times (I + I) = I$ and $I \times I + I \times I = I$, so $\forall x,y,z \in F, x(y+z) = xy+xz$

Based on the above, $\{I\}$ seems to qualify as a field.

If $I$ is assumed to be a real number, then the unique solution of $I + I = I$ and $I \times I = I$ is, of course, $I = 0$.

So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it? Specifically, is it required for the multiplicative identity to be distinct from the additive identity?

Answer 1

The short answer has already been given, that in general we do not want $1=0$. The element $I$ of the singleton set you are talking about is typically chosen to be $0$ instead, so that's why I'm using $\{0\}$ to denote the ring.

Here's one reason why the ring $\{0\}$ wouldn't make a good field:

$\{0\}^n\cong \{0\}$ for any $n$ as a "vector space," which doesn't fit very well with uniqueness of dimension for the vector spaces over other fields.

Although it is not about a bona-fide field, you might be interested in the wiki article on "field with one element". It gives motivation for an exceptional situation where it makes a little sense to think about something like a field with one element, but it isn't really a set satisfying the field axioms.

Again, the topic of the linked article in the previous paragraph is not actually a field at all, despite the name given to it. Seeing "field" and "one element" makes it tempting to conclude that $\{0\}$ is the subject of discussion (but it isn't).

Answer 2

One requires $1_{F}\neq0_{F}$ so $\{0\}$ is not a field.

Answer 3

I am quoting Lang's Algebra p. 84: "A ring $A$ such that $1 \neq 0$ and such that every nonzero element is invertible is called a division ring...A commutative division ring is called a field." Hence by definition, a field must contain at least two distinct elements.

Answer 4

In the comments I claimed that the zero ring isn't a field for the same reason that $1$ isn't prime. Let me make this connection precise.

By the Artin-Wedderburn theorem, any semisimple commutative ring is uniquely the direct product of a finite collection of fields (with the usual definition of field). This theorem fails if you allow the zero ring to be a field, since the zero ring is the identity for the direct product: then for any field $F$ we would have $F \cong F \times 0$.

The "correct" replacement for the axiom that every nonzero element has an inverse is the axiom that there are exactly two ideals. The zero ring doesn't satisfy this axiom because it has one ideal. This has geometric meaning: roughly speaking it implies that the spectrum consists of a point, whereas the spectrum of the zero ring is empty.

Analogously, if you want every graph to be uniquely the disjoint union of connected graphs, you need to require that the empty graph is not connected. Here the "correct" replacement for the axiom that there is a path between every pair of vertices is the axiom that there is exactly one connected component.

Answer 5

Let me compile some comments into an answer.

By now it seems clear that the question is predicated on a misstatement / faulty memory of the field axioms. I claim that it is completely standard that $1 \neq 0$ is included in the field axioms. By "completely standard" I mean that every reputable source on the subject in the last $50$ years does it this way, and that at least 99.9% of people who speak about "fields" in the abstract algebraic sense mean to include this axiom.

If you don't include this axiom, then, as others have pointed out, the only additional structure you get is what is usually referred to as the zero ring: the unique -- up to isomorphism -- ring with a single element, which therefore acts as both an additive identity $0$ and a multiplicative identity $1$.

(The zero ring is certainly a ring, although a small minority of reputable sources exclude it. They shouldn't do so: we want, for any ideal $I$ in a ring $R$, to define the quotient ring $R/I$. If $I = R$ then this quotient is the zero ring. You might think "Well, fine, we'll just say that we can't take the quotient by the ideal $I = R$. For instance, maybe we'll say that $R$ is not an ideal of $R$ at all." But this is a bad convention, especially the latter, since in practice an ideal is often given by a set of generators, and usually one cannot easily tell by looking at the generators whether they generate the unit ideal $R$ or not. For instance, if $a_1,\ldots,a_n$ are integers, then according to this convention we can only consider the ring $\mathbb{Z}/\langle a_1,\ldots,a_n \rangle$ if there is some $d \geq 2$ which divides each of $a_1,\ldots,a_n$. So we have work to do before we know whether we can write down the quotient ring or not! In more complicated rings this becomes more difficult or practically impossible.)

The reasons for excluding the zero ring from being a field have already been well-described by others. In particular the modern perspective on commutative rings is to focus attention on modules over the ring. (As an aside, the Morita theory in the commutative case simply says that two commutative rings with isomorphic module categories are isomorphic.) But the only module over the zero ring is the zero module. So unlike all other rings satisfying the remaining field axioms, there are no nontrivial vector spaces over the zero ring and hence no nontrivial linear algebra. We also have the important notion of a simple module, which is a nonzero module with only the zero module as a submodule, and there is the important fact that if $N \subset M$ are $R$-modules, then $N$ is a maximal submodule of $M$ iff $M/N$ is a simple module.

Then there is the notion of a simple commutative ring, which is a ring such that $R$ is simple as an $R$-module. This means that $R$ is not the zero ring but the only ideals of $R$ are $(0)$ and $R$. It is easy to see that a commutative ring if simple iff it is a field. In particular, since ideals are $R$-submodules of $R$, an ideal $I$ of $R$ is maximal iff $R/I$ is a field. As Qiaochu Yuan likes to say, the zero ring is "too simple to be simple": including it as a field would screw all this up.

Let me end with some comments about forgetting about the field axiom $1 \neq 0$. In the context of analysis courses, fields appear at the beginning but only in a very shallow way. Every analysis text I know which discusses fields includes $1 \neq 0$ among the field axioms. However, I just checked Rudin's Principles of Mathematical Analysis and Spivak's Calculus, and though they include it, they do so in a way which makes it easy to forget. Namely they put it in as a sort of rider in the axiom about the existence of a multiplicative identity. From the perspective of abstract algebra this is a strange approach, since the existence of a multiplicative identity is one of the ring axioms, and there is a ring in which $0 = 1$, so we shouldn't be saying that $0 \neq 1$ in the same breath as we are averring the existence of $1$. But you don't encounter the definition of a ring in real analysis, typically.

On the other hand, notice how the field axioms are stated in $\S$ 1.2 of these honors calculus notes written by an algebraist. It makes it harder to forget that $0 \neq 1$!