# Abstract Nonsense

## Semiring of Ideals

Point of Post: In this post we define the notion of a semiring and then define a semiring structure on the set $\mathfrak{L}(R)$ of ideals of a ring $R$.

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Motivation

In our last post we saw that every ring has associated to it a complete modular lattice $\left(\mathfrak{L}(R),\subseteq,\cap,+\right)$ which is comprised of its ideals. More particularly, we saw that there was a way to ‘add’ ideals together which was based off the addition of the ring itself. An obvious question then is whether or not we can figure out a way to ‘multiply’ ideals together which uses the multiplication of a ring. But then! We think “we’ve got a set with a multiplication and addition…that sounds like…well..you know“. So we pursue this and see if these new notions of addition and multiplication of ideals make $\mathfrak{L}(R)$ into a ring. Sadly, this isn’t the case. But, we get pretty darn close. We find out that $\mathfrak{L}(R)$ with these operations is a semiring which is roughly a ring where the additive structure is no longer an abelian group but a commutative monoid.

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Product of Ideals

Let $R$ be a ring. We define the multiplication of two ideals $\mathfrak{a},\mathfrak{b}\in\mathfrak{L}(R)$, denoted by concatenation, to be the ideal $\left(\left\{ab:a\in\mathfrak{a}\text{ and }b\in\mathfrak{b}\right\}\right)$. In other words, the product of two ideals is the ideal generated by the setwise product of the ideals. But, just as the case with the sum of ideals this formidable looking definition reduces very nicely. Namely:

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Theorem: Let $R$ be a ring and $\mathfrak{a,b}\in\mathfrak{L}(R)$. Then, $\mathfrak{ab}=\left\{a_1b_1+\cdots+a_nb_n:a_k\in\mathfrak{a},\;b_k\in\mathfrak{b}\text{ and }n\in\mathbb{N}\right\}$.

Proof: Denote the right hand side of the above equation as $J$. Evidently $J\in\mathfrak{L}(R)$ since it’s obviously a subgroup and left or right multiplying by a general $R$-element on a finite sum of the form $a_1b_1+\cdots+a_nb_n$ is just absorbed by the $a_k$‘s on the left and the $b_k$‘s on the right. Since it clearly contains the set $\left\{ab:a\in\mathfrak{a}\text{ and }b\in\mathfrak{b}\right\}$ we have that $\mathfrak{ab}\subseteq J$. Conversely, $J$ is evidently a subset of $\mathfrak{ab}$ since it contains all elements of the form $late ab$ with $a\in\mathfrak{a}$ and $b\in\mathfrak{b}$ and thus must contain all finite sums of that type. $\blacksquare$

$\blacksquare$

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In fact, from this it’s easy to see that if we define $\mathfrak{a}_1\cdots\mathfrak{a}_n$ to be equal to $\mathfrak{a}_1(\mathfrak{a}_2(\cdots(\mathfrak{a}_n)\cdots))$ then

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$\mathfrak{a}_1\cdots\mathfrak{a}_m=\left\{a_{1,1}\cdots a_{1,m}+\cdots a_{n,1}\cdots a_{n,m}:a_{i,j}\in\mathfrak{a}_j\right\}$.

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So an obvious question is how this product of ideals relates to former operations on ideals such as intersection and sum. Unsurprisingly, it distributes over addition but perhaps slightly unexpected (not really if you think about it but viscerally kind of weird) the product of ideals is contained in the intersection of the ideals. Indeed:

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Theorem: Let $R$ be a ring and $\mathfrak{a,b,c}\in\mathfrak{L}(R)$. Then:

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\begin{aligned}&\mathbf{(1)}\quad \mathfrak{a}0=0\mathfrak{a}=0\\ &\mathbf{(2)}\quad \mathfrak{a}(\mathfrak{b}+\mathfrak{c})=\mathfrak{ab}+\mathfrak{ac}\text{ and }(\mathfrak{b}+\mathfrak{c})\mathfrak{a}=\mathfrak{ba}+\mathfrak{bc}\\ &\mathbf{(3)}\quad \mathfrak{ab}\subseteq\mathfrak{a}\cap\mathfrak{b}\end{aligned}

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where we’ve used $0$ to denote the zero ideal $\{0\}$. Furthermore, if $R$ is commutative then

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$\mathbf{(4)}\quad(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b})\subseteq\mathfrak{ab}\quad\quad\text{ }\quad\quad\quad\;\text{ }_.$

Proof: $\mathbf{(1)}$ is obvious.

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To see $\mathbf{(2)}$ we let $x\in \mathfrak{a}(\mathfrak{b}+\mathfrak{c})$ then evidently $x=a_1(b_1+c_1)+\cdots+a_n(b_n+c_n)$ for some $a_k\in\mathfrak{a},b_k\in\mathfrak{b},c_k\in\mathfrak{c}$, and $n\in\mathbb{N}$. But, then $x=(a_1b_1+\cdots+a_nb_n)+(a_1c_1+\cdots+a_nc_n)$ and so $x\in \mathfrak{ab}+\mathfrak{ac}$. Conversely, we note that evidently $\mathfrak{ab},\mathfrak{ac}\subseteq \mathfrak{a}(\mathfrak{b}+\mathfrak{c})$ (since we can consider sums of the form $a(b+0)+\cdots$ and $a(0+c)+\cdots)$) and thus since it’s a subgroup we have that $\mathfrak{ab}+\mathfrak{ac}\subseteq\mathfrak{a}(\mathfrak{b}+\mathfrak{c})$. The other result is similar.

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To see $\mathbf{(3)}$ we merely note that if $a\in\mathfrak{a}$ and $b\in\mathfrak{b}$ then $ab\in\mathfrak{a}$ (by right absorption) and $ab\in\mathfrak{b}$ (by left absorption) so that $ab\in\mathfrak{a}\cap\mathfrak{b}$. Thus, $\mathfrak{a}\cap\mathfrak{b}$ is an ideal of $R$ containing the set of all $R$-elements of the form $ab$ so that, by definition, $\mathfrak{ab}\subseteq\mathfrak{a}\cap\mathfrak{b}$.

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To see $\mathbf{(4)}$ we merely note that if $x\in(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b})$ then $x=(a+b)c$ where $a\in\mathfrak{a}$, $b\in\mathfrak{b}$, and $c\in\mathfrak{a}\cap\mathfrak{b}$. But then, $x=ac+bc=ac+cb$ but evidently $ac,cb\in\mathfrak{ab}$ so $ac+cb=x\in\mathfrak{ab}$. Since $x$ was arbitrary the conclusion follows. $\blacksquare$

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Semiring of Ideals

We now show the true extent of the interaction between the product and sum of ideals. To put it in the right light, we give the ‘structure’ $\mathfrak{L}(R)$ with the sum and product of ideals takes on a name. Namely, a set $(S,+,\cdot)$ is called a semiring if $(S,+)$ is a commutative monoid with identity $0$, $(S,\cdot)$ is a semiring, $+$ left and right distributes over $+$, and finally $a\cdot 0=0\cdot a=0$. What we now claim is that $\mathfrak{L}(R)$ with the sum and product of ideals forms a semiring. Indeed:

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Theorem: Let $R$ be a ring, then $\left(\mathfrak{L}(R),+,\cdot\right)$ forms a semiring with additive identity $0=\{0\}$.

Proof: To see that $(\mathfrak{L}(R),+)$ is a commutative monoid is easy since evidently the addition of ideals inherits associativity from the associativity of the $R$-addition, it similarly inherits the commutativity from the $R$-addition, and evidently $\mathfrak{a}+0=0+\mathfrak{a}=\mathfrak{a}$.

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It’s easy to see that $(\mathfrak{L}(R),\cdot)$ is a semiring since the product of ideals inherits associativity from the $R$-product.

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The last two properties, left and right distributivity as well as left and right annhilation by $0$ were proven in the last theorem. $\blacksquare$

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

June 23, 2011 -

## 3 Comments »

1. […] also fairly clear to see that is a semiring  homomorphism. Moreover, if I would have defined the lattice of subrings we would see that the […]

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2. […] implies or . Or, taking it one step further the equivalent definition that and (recalling the product of ideals) implies or . Thus, we are lead to the notion of prime ideals in a general ring which are ideals […]

Pingback by Prime Ideals (Pt. I) « Abstract Nonsense | August 17, 2011 | Reply

3. […] true for , and so suppose it’s true for and let are ideals of We see then that . Recall that, in general we have the relations and . But, by the previous paragraph we have that so that […]

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