Is there a field (or field-like structure) that is distributive over both operations?

Question

Is there a field (or field-like structure) that is distributive over both operations?

i.e. $(F,+,*)$ such that $\forall a,b,c \in F $

1. $a+(b*c) = (a+b)*(a+c) $
2. $a*(b+c) = (a*b)+(a*c)$

P.S.: For example, conjuction and disjunction both satisfy these properties.

P.S.S:
Field axioms:

1. $(F,+)$ is an abelian group.

2. Multiplication is associative

3. $\exists 1\in F$ such that $\forall a\in F$, $a*1 = 1*a$.

4. $\forall a\in F$, if $a\neq 0$ then there exists $b\in F$ such that $a*b=b*a = 1$.

5. Multiplication is distributive over addition

Answer 1

If $(F,+,*)$ is a structure in which $(F,+)$ is an abelian group with identity $0$, $(F,*)$ is a commutative monoid with identity $1$ (your axioms 1, 2, and 3); $*$ distributes over $+$ (your axiom 5); and $+$ distributes over $*$, then you just get a structure with a single element $\{0\}$. We don't even need multiplicative inverses.

Indeed, first, note that $a*0 = 0$ for all $a$, since $a*0 = a*(0+0) = (a*0)+(a*0)$, and cancelling we get $a*0 = 0$.

Next, note that $a^2=a$ for all $a$. Indeed, since $+$ distributes over $*$, we have $$a = a+0 = a+(0*0) = (a+0)*(a+0) = a*a = a^2.$$

And if $1$ is the identity element of $*$, then since $+$ distributes over $*$ and $*$ distributes over $+$, we have: $$\begin{align*} a &= a+0\\ &= a+(1*0)\\ &= (a+1)*(a+0)\\ &= (a+1)*a \\ &= a^2+(1*a)\\ &=a^2+a\\ &=a+a. \end{align*}$$

But then, because $(F,+)$ is an abelian group, we can cancel $a$ from $a=a+a$ to get $a=0$.