Is $\{0\}$ a field?

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?

Solution 1:

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.

Solution 2:

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).