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Let $S$ be a commutative semiring with identity such that each prime ideal of $S$ is subtractive. Does this imply all ideals of $S$ to be subtractive?

By a commutative semiring with identity I mean an algebraic structure, consisting of a nonempty set $S$ with two operations of addition and multiplication such that the following conditions are satisfied:

$(S,+)$ is a commutative monoid with identity element $0$; $(S,.)$ is a commutative monoid with identity element $1 \not= 0$; Multiplication distributes over addition, i.e. $a(b+c) = ab + ac$ for all $a,b,c \in S$; The element $0$ is the absorbing element of the multiplication, i.e. $s.0=0$ for all $s\in S$.

A nonempty subset $I$ of a semiring $S$ is said to be an ideal of $S$, if $a+b \in I$ for all $a,b \in I$ and $sa \in I$ for all $s \in S$ and $a \in I$.

A nonempty subset $P$ of a semiring $S$ is said to be a prime ideal of $S$, if $P \not= S$ is an ideal of $S$ such that $ab \in P$ implies either $a\in P$ or $b\in P$ for all $a,b \in S$.

An ideal $I$ of a semiring $S$ is said to be subtractive, if $a+b \in I$ and $a \in I$ implies $b \in I$ for all $a,b \in S$.

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I think the answer is no. Let $S=\{0\}\cup[1,\infty)$ be the subsemiring of the (usual) reals. A non-zero ideal of $S$ is of the form $\{0\}\cup[a,\infty)$ where $a\geq 1$. Clearly, the only prime ideal of $S$ (according to your definition) is $\{0\}$ and it is subtractive. But no proper non-zero ideal of $S$ is subtractive.

Correction: My argument is not right: actually every non-zero ideal of $S$ ie either of the form $\{0\}\cup[a,\infty)$ or $\{0\}\cup(a,\infty)$ where $a\geq 1$. Hence $P=\{0\}\cup(1,\infty)$ is a non-zero prime ideal which is not subtractive. And in general a unitary semiring has to always have a maximal ideal (as the unitary ring does.)

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