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Do finite non-associative fields exist? By “non-associative field”, I mean a field with associativity of multiplication dropped but all other properties remain the same.

This other question asks something similar but does not require commutativity of multiplication.

Update: according to A089925, any non-associative field must have at least 7 elements (one more than is needed for a non-associative Abelian loop due to having a zero element).

Super Cool Guy
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Daniel Sebald
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    In about 12 hours, your question got upvoted two times, yet it has been closed without any comment. Why? – Dominique Nov 06 '25 at 15:57
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    I second this. The question is clear enough: the terminology of a "non-associative field" is non-standard, but is defined within the question itself. – ViHdzP Nov 27 '25 at 21:41

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Without commutativity requirement, these structures are called semifields, so we are looking for commutative semifields. There is a survey of the constructions: M.Cordero and G.P.Wene "A survey of finite semifields", Discrete Math 208/209:125-137 DOI 10.1016/S0012-365X(99)00068-0.

The minimal size commutative semifield has 27 elements. It is a nonassociative algebra of dimension 3 over the field with 3 elements, generated by 1 element, with the multiplication table given by

$1$ $a$ $aa$
$1$ $1$ $a$ $aa$
$a$ $a$ $aa$ $1+a$
$aa$ $aa$ $1+a$ $2+a+aa$
Vladimir Lysikov
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