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In a vector space $V$ over a field $F$, a nonempty subset $W$ of $V$ is a subspace if it is closed under addition and scalar multiplication. For a module $M$ over a ring $R$ with identity the similar result is true. Is it true in modules over a nonunital ring?

I should mention that the analogue of the following is not true.

In a vector space $V$ over a field $F$, a nonempty subset $W$ of $V$ is a subspace if for all scalars $\alpha_1, \alpha_2$ and for all $w_1, w_2 \in W$, $\alpha_1 w_1 + \alpha_2 w_2 \in W$.

This can be seen by the module $\mathbb{Z}$ over $2\mathbb{Z}$ and the subset $2\mathbb{Z} \cup \{-3,3\}$.

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In the category of models in Set of an algebraic theory, any subset that is closed under the operations of that theory is a subalgebra. For modules over a nonunital ring, the operations consist of the abelian group structure plus the ring actions. If your ring is unital then the unary operations of the abelian group structure are contained within the ring structure, if nonunital then you need to add them in explicitly. Essentially, you need to add the unit and then test for closure. –  Andrew Stacey Feb 11 '10 at 8:25
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This has been retagged several times! Please justify any further retagging in the "notes on edit" field. –  Andrew Stacey Feb 11 '10 at 10:28
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2 Answers

up vote 2 down vote accepted

In Hungerford everything is defined without assuming that rings have identity (or that they are commutative). At least according to Definition IV.1.3 on p. 171 of Hungerford, the answer to your first question is affirmative. Quoting:

Definition 1.3. Let $R$ be a ring, $A$ an $R$-module and $B$ a nonempty subset of $A$. $B$ is a submodule of $A$ provided that $B$ is an additive subgroup of $A$ and $rb\in B$ for all $r\in R$, $b\in B$.

[I hope that my interpretation of the question was correct. I understood 'characterization of submodule' as 'definition of submodule.']

EDIT: Google books links to the definition of a module, a submodule, and a ring in Hungerford:

http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA169#v=onepage&q=&f=false

http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA171#v=onepage&q=&f=false

http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA115#v=onepage&q=&f=false

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Are you sure about that? If I remembered correctly, Hungerford made such assumptions that R is unital or at least R^2=R –  Shizhuo Zhang Feb 11 '10 at 14:08
    
Quite sure: In Definition IV.1.1 on p. 169 (definition of a module) there is a part saying "If $R$ has an identity element... then $A$ is said to be a unitary $R$-module." But please check this in Hungerford to get convinced, perhaps I am missing something. –  user2734 Feb 11 '10 at 14:23
    
look, here. It is unitary R-module. What we are talking about is module over non-unital ring –  Shizhuo Zhang Feb 11 '10 at 15:33
    
Isn't a non-unital ring just a ring without a multiplicative identity? If the answer is yes, then at least as far as I can see, in Hungerford there is no assumption that rings have identity in the aboe definition of a submodule. But if I am totally missing something here, please let me know... –  user2734 Feb 11 '10 at 15:46
    
non-unital ring is indeed a ring without a multiplicative identity. When he defined a module, he considered the module over unital ring. So the other definitions should follow this assumptions –  Shizhuo Zhang Feb 11 '10 at 16:16
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First, Let me talk about the "correct definition" of module over non-unital ring(not necessarily commutative) and how this definition coincide with usual definition of module over unital ring in particular case

First we study $R-mod_{1}$={category of associative action of $R$ on $k$-mod}= {($M$,$R\bigotimes _{k}M\rightarrow M$).

$r_{1}(r_{2}z)=(r_{1}r_{2})z$}

Let $R_{1}=R\bigoplus k$ be an untial $k$-algebra with usual multiplication. And we have the categorical equivalence as: $R-mod_{1}\approx R_{1}-mod$

Now,we define module over non-unital algebra $R$ as $R-mod=R_{1}-mod/(Tors_{R_{1}})^{-}$, where $(Tors_{R_{1}})^{-}$ is Serre subcategory of $R_{1}-mod$

$R_{1}-mod\overset{q_{R}^{*}}{\rightarrow}R-mod$ is a localization functor having right adjoint functor.

Trivial Example:

if $R$ has is an unital $k$-algebra. Then $R_{1}-mod$ is equivalent to $R-mod$

Less Trivial example in commutative case:

Consider affine line $k[x]$. Let $R=xk[x]$(maximai ideal of $k[x]$). Then $R-mod$=$Qcoh(\mathbb{A}^{1}-{0}$). It is a cone.

Toy general case:

Let $m$ is a two-sided proper ideal of associative commutative unital ring $A$. Then: we have

$m-mod$=$A-mod/({M\epsilon A-mod|m\cdot M=0})^{-}$, where$T^{-}$ is smallest Serre category containing $T$. It is clear that is equivalent to Qcoh(Complement of $\mathbb{V}(m)$),where $\mathbb{V}(m)$ is closed subvariety determined by $m$.

Now, I should stop here and write another(maybe)post on definition of sub-module. There are several reference:

Gabriel, Pierre Des catégories abéliennes. (French) Bull. Soc. Math. France 90 1962 323--448 Kontsevich-Rosenberg Noncommutative spaces and flat descent

Gabber-RameroAlmost Ring Theory

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Interesting and wonderful as the general theory might be: what relevance does this have to the original question? It seems to me that an answer should take into account the level which the questioner either desires or will find helpful. QCoh strikes me as rather more high-powered than the original post was looking for... –  Yemon Choi Feb 12 '10 at 8:33
    
Hi Yemon. Actually, if one consider submodule of a module over this non-unital ring $R$, $M\right arrow N$, where $N\in R-mod$ and $M\rightarrow N$ is monomorphism, then $N$ is submodule of $M$. And I almost can say for sure that the naive definition of the module over non-untial ring is not correct –  Shizhuo Zhang Feb 20 '10 at 7:59
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