Difference between revisions of "Zero ring"

(this is slightly less silly than the zero module)
 
(Fixed a false statement that 0 is initial and not final; other small corrections/elaborations.)
 
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<cmath> x = 1 \cdot x = 0 \cdot x = 0 . \qquad \blacksquare </cmath>
 
<cmath> x = 1 \cdot x = 0 \cdot x = 0 . \qquad \blacksquare </cmath>
  
In the [[category]] of [[ring]]s, the zero ring is an [[initial object]].
+
In the [[category]] of [[ring]]s, the zero ring is a [[terminal object]], through the trivial ring homomorphism.
However, it is ''not'' a [[terminal object]]: in fact, the only rings
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However, it is ''not'' an [[initial object]].  This can be seen by the fact that ring homomorphisms must preserve the identities.  Clearly a ring homomorphism, <math>\varphi(1_0)</math> where <math>\varphi : \mathbf{0} \to R</math> cannot be defined as both <math>1_R</math> and <math>0_R</math> since AoPS defines rings to have multiplicative identity (some sources vary here).  It can instead be shown that the [[integers]], <math>\mathbb{Z}</math> is initial in <math>\mathbf{Ring}</math>.
with homomorphisms into the trivial rings are themselves trivial, as
 
we require [[ring homomorphism]]s to preserve 0 and 1.
 
  
 
Note that by convention there is no "trivial field" or "zero field",
 
Note that by convention there is no "trivial field" or "zero field",

Latest revision as of 17:44, 31 January 2022

A zero ring is a ring with one element, 0 (equal to 1), with the additive and multiplicative structure of the trivial group. Technically speaking, there are infinitely many zero rings (one for each possible element "0"), but they are all trivially isomorphic, so by abuse of language we may refer to the zero ring.

Proposition. If $R$ is a ring in which $0=1$, then $R$ is a trivial ring.

Proof. For any $x \in R$, we have \[x = 1 \cdot x = 0 \cdot x = 0 . \qquad \blacksquare\]

In the category of rings, the zero ring is a terminal object, through the trivial ring homomorphism. However, it is not an initial object. This can be seen by the fact that ring homomorphisms must preserve the identities. Clearly a ring homomorphism, $\varphi(1_0)$ where $\varphi : \mathbf{0} \to R$ cannot be defined as both $1_R$ and $0_R$ since AoPS defines rings to have multiplicative identity (some sources vary here). It can instead be shown that the integers, $\mathbb{Z}$ is initial in $\mathbf{Ring}$.

Note that by convention there is no "trivial field" or "zero field", as we usually require 0 and 1 to be distinct in fields.

See also