This is a scratchpad for experts in commutative algebra and models of linear logic to put together a possible solution of this problem.

It is probably best to take the ground field $K$ to be $\mathbb C$.

I suspect that Galois theory and insisting on fields are not going to help.

Also, this is not an undergraduate problem.

By linear algebra is meant the category $\mathbf V$ of vector spaces and linear maps over $K$, but it is likely that these will have to be topological.

In this category,

• the initial object $\bf 0$ is the zero vector space,
• the terminal object $\bf 1$ is also the zero vector space,
• the additive sum $A+B$ is the coproduct, which is the direct sum of vector spaces,
• the additive product $A\times B$ is the product, which is also the direct sum of vector spaces,
• the multiplicative unit $I$ is the ground field $K$,
• the multiplicative product $A\otimes B$ is the tensor product over $K$, and
• the linear hom $A\multimap B$ is the hom-set ${\bf A}(A,B)$ together with the vector space structure inherited from $B$.

There is likely to be some debate over the following:

• there is a dualising object $\bot$ that may also be the ground field $K$
• the dual $A^\bot$ is $A\multimap\bot$, but this need not be involutive ($(A^\bot)^\bot=A$),
• the par (using Girard's name) $A @ B$ must satisfy $(A @ B)^\bot=(A^\bot)\otimes(B^\bot)$ and may also be the tensor product, but need not be.

By algebras is meant the category $\mathbf A$ of commutative unital rings and homomorphisms over this field, but it is likely that these will have to be topological.

Often in models of linear logic the functor $? A$ (why not in Girard) is the free monoid on $A$, ie the free algebra generated by $A$ qua vector space. The functor $! A$ satisfies $!(A^\bot)=(? A)^\bot$.

However, there is a great deal of freedom in the choice of these operations. There can be models with the same underlying linear algebra but different $!$ operations.

We do need Seely's equations: $!(A\times B)=(!A\times !B)$ and $!\mathbf{1}=I$.

As a first step we also need to identify $\Sigma=!I$ and $R=!\Sigma$.

Very likely, $\Sigma=K[x]$, the polynomial ring in one variable.

$R$, considered as a variety, has as points the elements of the ground field. (Please would some commutative algebraist replace this with a more precise description.)