most recent 30 from http://mathoverflow.net 2013-06-19T01:24:26Z http://mathoverflow.net/feeds/question/105026 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105026/semirings-with-subtractive-primes Peyman Nasehpour 2012-08-19T07:05:18Z 2012-12-08T16:08:11Z

<p>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?</p> <p>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:</p> <p>$(S,+)$ 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 $s\in S$.</p> <p>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 $a \in I$.</p> <p>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 $a,b \in S$.</p> <p>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 $a,b \in S$.</p>

http://mathoverflow.net/questions/105026/semirings-with-subtractive-primes/115756#115756 Miroslav Korbelar 2012-12-07T23:08:15Z 2012-12-08T16:08:11Z

<p>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.</p> <p>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.) </p>