2
$\begingroup$

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 $a,b \in S$.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.