# 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?

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 $F \cong F \times 0$.