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.
In particular: 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 $\mathsf{Pr} \infty \mathsf{fWiki}$ 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

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