This is my solution of Problem 2, part (a). Please let me know if there are any mistakes.
It was proven in Theorem 3.6 of the paper "A Golbach conjecture for Laurent Series Semidomain" that if is an additively reduced and additively atomic semidomain, the following conditions are equivalent
(1)
(2) every with |supp can be expressed as the sum of at most 2 irreducibles.
Consider the additive monoid . It is well know that is atomic and . Since is also closed with respect to the usual multiplication of rational numbers, we know that is a semidomain. Now consider the multiplicative monoid . It is clear that is a multiplicative subset of , and then we can consider the localization .
Claim 1: We see that .
Proof: Let us consider an element . This means that there are and such that . Take for some , and we can suppose that for some . Note that then
Since for every , we can conclude that .
Suppose now that . We can write then for some , and some . For each we can find some natural numbers and such that . Then we can write which is clearly an element of .
Let us denote . In the paper "Puiseux monoids and transfer isomorphism", Dr. Felix Gotti proved in Proposition 3.5 that if , then the additive monoid is atomic with . Then, by the theorem mentioned above, we know that the semidomain satisfies Goldbach conjecture. However clearly does not satisfy it. Indeed, for every the polynomial cannot be expressed as the sum of or fewer irreducibles.
One of the irreducible criteria is well-known for polynomials with integers coefficients and also holds naturally for polynomials of is Eisenstein's irreducible criterion that states
Criterion 1 (Eisenstein's irreducible criterion).
Let . If there exists a prime number satisfying the following three conditions:
- divides each for . - does not divide , and - does not divide .
Then the polynomial is irreducible.
We have another criterion, which is even easier to check.
Criterion 2: Let . Suppose that and is a prime number, then is irreducible.
Proof
Indeed, suppose that for some . Note that , and since is prime, we know that either or . WLOG, assume that , then for some . Note that if , then , and then , which contradicts the hypothesis. Therefore , and this implies that is irreducible.
First, suppose that . Note that we can find a polynomial of that is not irreducible. We can consider, for example, . Fix and consider the polynomial . Suppose towards a contradiction that with . Note that we have at least one such that . Indeed, if for every , then the constant coefficient of is at least . However if , then we have that , which is a contradiction with the irreducibility of . This proves the result.