# Smallest Field is Field

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## Theorem

The ring $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest algebraic structure which is a field.

## Proof

From Field Contains at least 2 Elements, a field must contain at least two elements.

Hence the null ring, which contains one element, is *not* a field.

For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a field:

- $\struct {\set {0_R, 1_R}, +}$ must be an abelian group.

This is fulfilled as this is the parity group.

- $\struct {\set {0_R, 1_R}, \circ}$ must be a commutative division ring.

This is fulfilled, as $\struct {\set {0_R, 1_R}^*, \circ} = \struct {\set {1_R}, \circ}$ is the trivial group.

- $\circ$ needs to distribute over $+$. This follows directly from Ring Product with Zero and the behaviour of the identity element in a group.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $1$: Integral Domains: $\S 5$. Further Examples of Integral Domains: Example $6$ (which proves the long way round that this is an integral domain) - 1973: C.R.J. Clapham:
*Introduction to Mathematical Analysis*... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $4$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 2.4$: The rational numbers and some finite fields