Smallest Field is Field
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.
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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