# Smallest Field is Field

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Although this article appears correct, it's inelegant. There has to be a better way of doing it.Amend this, or add another proof, to use the ring of integers modulo $2$ or something. It's currently unwieldy. We need to show a) that this is a field, and b) that this is the smallest field.You can help Proof Wiki by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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