Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is related to this one. This will be quite a long post. Section "Introduction" may be skipped at first. I will refer to it later but it may be easier to only skim that section first (especially given that it's common knowledge).

Preliminary definitons

A semiring is a set with two operations $+$ and $\cdot$ which satisfy all axioms of a unitary ring except the axiom demanding the existence of additive inverses.

A left semimodule over a semiring $R$ is a commutative monoid $M$ (whose operation is denoted by $+$ too) with a function $ R\times M\longrightarrow M$ satisfying all axioms of a module over a ring except the axiom demanding the existence of additive inverses. A right semimodule is defined analogously.

Introduction

There are two equivalent definitions of a basis of a left $R$-module $M$ for a nontrivial unitary ring $R$.

Definition 1. A set $X\subset M$ is a basis of $M$ when $X$ is linearly independent (that is a linear combination of elements of $X$ is zero iff its coefficients are all zero), and $X$ generates $M$.

Definition 2. A set $X\subset M$ is a basis of $M$ when for every left $R$-module $N$ and every function $f:X\longrightarrow N$ there exists a unique $R$-module homomorphism $\bar f:M\longrightarrow N$ such that $\bar f|_X=f.$

I will prove the equivalence of these definitions now so that I can show where I have problems with the analogous defintions for semimodules.

Proof. Suppose $X$ satisfies Definition 1. Let $N$ be a left $R$-module and $f:X\longrightarrow N.$ We define $\bar f:M\longrightarrow N$ by $$\bar f\left(\sum_{i=1}^nr_ix_i\right)=\sum_{i=1}^nr_if(x_i).$$

Since $X$ generates $M$, this defines $\bar f$ for every $m\in M$. We have to show that it is well defined. Suppose $$\sum_{i=1}^nr_ix_i=\sum_{i=1}^ns_ix_i.$$ Then, by the linear independence of $X$, $$r_i=s_i$$ for all $i=1,\ldots,n.$ Therefore,

$$\bar f\left(\sum_{i=1}^nr_ix_i\right)=\sum_{i=1}^nr_if(x_i)=\sum_{i=1}^ns_if(x_i)=\bar f\left(\sum_{i=1}^ns_ix_i\right).$$ We have proved the existence of $\bar f$. We need to show its uniqueness. Suppose $g:M\longrightarrow N$ is a homomorphism, and $g|_X=f$. Then

$$g\left(\sum_{i=1}^nr_ix_i\right)=\sum_{i=1}^nr_ig(x_i)=\sum_{i=1}^nr_if(x_i)=\bar f\left(\sum_{i=1}^nr_ix_i\right),$$ and since $X$ generates $M,$ this proves the uniqueness of $\bar f.$

Conversely, suppose $X$ satisfies Definition 2. We will first prove that $X$ is linearly independent. Suppose $$\sum_{i=1}^nr_ix_i=0,$$ and $r_1\neq 0$. Let $f:X\longrightarrow R$ be defined by $$f(x)=\begin{cases}1 & \text {for } x=x_1,\\0 & \text {otherwise.}\end{cases}$$

Then there exists an $R$-module homomorphism $\bar f:M\longrightarrow R$ such that $\bar f|_X=f.$ We have $$0=\bar f(0)=\bar f\left(\sum_{i=1}^nr_ix_i\right)=\sum_{i=1}^nr_if(x_i)=r_1.$$ This is a contradiction, and $X$ must be linearly independent. We need to show that $X$ generates $M$. Let $M_1\subseteq M$ be the submodule generated by $X.$ Let $\bar f:M\longrightarrow M/M_1$ be the canonical homomorphism. Clearly, for all $x\in X,$ we have $\bar f(x)=0.$ Then $\bar f$ and the zero homomorphism restricted to $X$ are equal, and must therefore be equal. Hence $M/M_1=0,$ and so $M=M_1,$ which ends the proof. $\square$

Question

As said in the question I linked to at the beginning, the definition of linear independence in semimodules requires a bit of tweaking. If I understand correctly what Bill Cook wrote in his answer and what I later read in a book on semirings, the correct definition is this:

Definition 3. For a semiring $R$ and a left $R$-semimodule $M,$ a set $X\subset M$ is linearly independent iff for any two linear combinations of elements of $X$, their equality implies the equality of their coefficients.

This gives us two possible definitions of a basis of a left $R$-semimodule $M$ over a semiring $R$ obtained by re-writing Definition 1 and Definition 2, but now considering $M$ to be a semimodule over a semiring $R$. Let's call these definitions Definition 1s and Definition 2s.

Are these definitions also equivalent? I think they should be. Definition 1s is part of the definition of a free semimodule given by J. S. Golan in Semirings and Their Applications. But a definition of a free semimodule using Definition 2s should also be possible. However, it's difficult for me to generalize the proof I gave above. Most of it generalizes easily. I will not write those parts again, now for semimodules, but it is easy to check that Definition 1s implies Definition 2s by imitating the proof above. It also allows one to prove that Definition 2s implies the linear independence of $X$. What doesn't seem to be working is the proof that Definition 2s implies that $X$ generates $M$. The problem is that quotient structures are not very nice for semimodules. When we quotient out a submodule of a module over a ring, we use subtraction implicitly -- $m, m'$ being in the same coset means that $m-m'\in M_1.$ Quotient semimodules are generally defined not by subsemimodules, but by congruences. (As it is the case for semigroups.)

I've found a definition of a quotient semimodule $M/N$ for a subsemimodule $N\subseteq M$, but I don't think it works here. It is defined by the so-called Bourne relation.

Definition 4. Let $R$ be a semiring and $M$ a left $R$-semimodule. Let $N\subseteq M$ be a subsemimodule. We define the relation $\equiv_N$ on $M$ by $$m\equiv_N m'\iff (\exists n,n'\in N)\; m+n=m'+n'.$$

This relation can be shown to be a congruence. Now we simply define $M/N$ as $M/\equiv_N.$ This is easily seen to be equivalent to the standard definition in the case of modules over a ring.

However, as far as I can see, under this definition, $M/N=0$ doesn't imply $M=N$ in the general case. And this is what we want to use...

share|improve this question
    
I was just thinking that with $M=\{z\in\mathbb{Z}\mid z\geq 0\}$ and $N=\{z\in\mathbb{Z}\mid z\geq 2\}\cup\{0\}$ would produce $M/N=0$. I have another thing to add. One of Golan's main theses is that "semirings are not the poor man's ring". That is to say, looking at them as if they were bad rings is not the way to go about it. I find lattice ordered semirings to be a bit more interesting, because they are modeled on the lattice of ideals of a ring. There is a whole theory of lattice quotients an localizations there that may be more comfortable. –  rschwieb Jun 5 '12 at 16:44
    
Thanks for the example. When I noticed that the simple way of proving $M/N=0\implies M=N$ doesn't work, I was so sure it couldn't be true that I didn't even think about an example. I don't have the book at hand at the moment, so I can't find his explanation of this statement, but I had the opposite impression when I was reading. Most of the important definitions there seemed to be taken directly from ring theory with some changes when necessary. Of course there were other definitions: there are no additively idempotent rings. But ideals, modules and zero divisors are all there. –  user23211 Jun 5 '12 at 21:30
    
@rschwieb Anyway, I think one doesn't need to assume that semirings are worse rings to ask my question. My Definition 1<sup>s</sup> is Golan's too, and Definition 2<sup>s</sup> is a category-theoretic one, not ring-theoretic, I think. –  user23211 Jun 5 '12 at 21:32
    
He emphasizes this theme more in his article Semirings for the ring theorist, which I recommend. I'm not attacking the philosophy of the question (I've really liked every question I've read of yours :) ), I just wanted to float the idea that ideas about rings often fall through with semirings. One example Golan gives is of the Jacobson radical of a semiring. From the way he talks about it, it has no value for semirings. Judging from the categorical and elementary importance of the JR for ordinary rings, this is quite a surprise! –  rschwieb Jun 5 '12 at 21:33
    
@rschwieb Thanks! I definitely want to see it. The book doesn't focus on explaining the ideas behind the various definitions and theorems. Sometimes it's a bit like a list of theorems. –  user23211 Jun 5 '12 at 21:42

2 Answers 2

up vote 2 down vote accepted
+250

I think we gave up on the "normal" definition of a quotient semimodule too early, and then got distracted with the Bourne relation.

We can't write $r\sim s\iff r-s\in N$ as you said, but we can write $r\sim s \iff r+N=s+N$.

This resembles the Bourne relation written above, but of course the quantifiers are different. This equality of sets allows much more flexibility, and not just the existence of two things in $N$ "lining up" $r$ and $s$.

Now if $M/N=0$ in this relation, and $M$ and $N$ contain additive identity $0$ (which I presume was intended to be kept in the definition of semimodule above) then $m\sim 0$ implies $m+N=0+N$ for all $m\in M$, and so in particular $m+0\in N$, i.e. $m\in N$. Hence, $M\subseteq N$.

So I think your original argument does work.

share|improve this answer
    
Thanks. Is this relation always a congruence? That is, if $t\in R$, does $r+N=s+N$ imply $tr+N=ts+N$? I can't see why it should. I tried to post a counterexample here before, but it was completely wrong so I deleted it. Still thinking though. –  user23211 Jun 9 '12 at 8:31
    
@ymar $r+N=s+N\implies r=s+n\implies tr=ts+tn\in ts+N$, so $tr+N\subseteq ts+N$, and a symmetric argument proves the other containment... that's what I thought anyway. –  rschwieb Jun 9 '12 at 11:16
    
If M is a semimodule and N a subsemimodule of M. What is the structure of m+N in semimodule category? –  user33751 Jun 9 '12 at 13:27
    
m+N for every m in M –  user33751 Jun 9 '12 at 13:28
2  
@ymar Just like for group quotients, you need the subobject to be "normal" for the quotient to have structure. Normal here means, of course, $a+N=N+a$ for all $a$. Commutativity of addition is a convenient way to ensure all subsemigroups are normal. With this property, the obvious operation $(a+N)+(b+N)=(a+b)+N$ is well defined. –  rschwieb Jun 9 '12 at 23:49

There is a different proof working for modules and semimodules. I show only the generating property from Definition 2.

Let $X$ satisfy Definition 2. Let $M_1$ be the subsemimodule of $M$ generated by $X$. Let $i: X \to M_1$ and $j: M_1 \to M$ be the embeddings. Use $f = i: X \to M_1$. Then there is a unique $\bar f: M \to M_1$ with $\bar fji = f$. Because of $j\bar fji = jf = ji = id_M ji$ we get $j\bar f = id_M$ by the uniqueness requirement. Hence the embedding $j$ is surjective and thus the identity.

Here is an interesting example (theorem) w.r.t. $M/N = 0$. I am using the congruence relation in Definition 4.

Thm: Let $N$ be a subsemimodule of $N_0$, the natural numbers. Then $N_0/N = 0$ iff there are $a, b$ in $N$ with $a = b+1$. (1 not necessarily in $N$.)

There are infinitely many such subsemimodules in $N_0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.