An idempotent semifield is an idempotent semiring that has a zero and in which every non-zero element is invertible. Or, in other words, a skewfield in which an element may not have a negative but is always idempotent with respect to addition.
An idempotent semifield $K$ is idempotent semiring that has a zero $0$ and in which for every non-zero element $x$ there is an element $y$ such that $x \cdot y = 1$. It is said to be commutative if the multiplication is commutative.
As in the case of an idempotent semiring there is a partial order given by
which has a join given by addition. In case of an idempotent semifield this partial order also has a meet. To each semifield there is a dual semifield $K^*$ given on the same set of elements as $K$ and with the same zero, identity, and multiplication but with addition given by interchanging join and meet, i.e. the meet becomes the new addition on non-zero elements and the zero element behaves neutral.
A version of the fundamental theorem of algebra can be formulated for semifields: A semifield is said to be algebraically closed if the equation $x^n = y$ has a solution for all $x\in K$ and $n=1,2,\ldots$.
In an algebraically closed commutative idempotent semifield $K$ with $0 \cdot a = 0$ for all $a \in K$ the equation
with $y,p_0, \ldots, p_n \in K$ and $n=1,2,\ldots$ has a solution in $K$ if and only if $y \geq p_0$. If $p_0 = 0$, then the solution is unique and can be expressed in terms of the coefficients with help of radicals, multiplication, inverse and meet (i.e. a polynomial in the radicals of $y, p_0, \ldots, p_n$ over the dual semifield).
The max-plus algebra $(-\infty, +\infty]$ with addition given by maximum and multiplication given by ordinary addition.
The two element semifield $\{ 0, 1 \}$ with $0 \leq 1$ and $+ = \cdot = \vee$ is commutative and algebraically closed but does not fulfill the assumptions of theorem . Indeed, the equation $1 \cdot x = 1$ does not have a unique solution.
In the reference the assumption that $0 \cdot a = 0$ is not explicit (otherwise the base step in the induction in the proof of Proposition 2 therein doesn’t work).
Last revised on February 25, 2019 at 06:39:26. See the history of this page for a list of all contributions to it.