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I just started learning about modules. What happens if a free module is generated using elements of a finite set $X$ as a basis, but by multiplying each basis with a different ring? For example, let $X = \{a,b\}$, and consider, $$ M = a\mathbb{Z}_3 + b\mathbb{Z}_5. $$ Is this a module? A sum of modules? Another object?

If I take, $$ N = (b-a)\mathbb{Z}_3. $$ Can I say what M / N is? Would it be isomorphic to $a(\mathbb{Z}_3 + \mathbb{Z}_5)$?

Thank you very, very much.

Edit 1

From what the others say, who know much more than me, the setup does not make sense with $\mathbb{Z}_3$, $\mathbb{Z}_5$. What if I consider the following version instead?

$$ M = a\mathbb{R}[x] + b\mathbb{R}[y]. $$ Is this a module? A sum of modules? Another object?

If I take, $$ N = (b-a), $$ Can I say what M / N is? Would it be isomorphic to $a(\mathbb{R}[x] + \mathbb{R}[y])$?

Edit 2

I just found this that seems relevant: modules over direct sum of different rings

r1697464
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    What is $a(\mathbb{Z}/n)$, if $a$ is not an integer? – Dietrich Burde Nov 02 '17 at 13:46
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    To define a module you would have to define one ring over which is module is defined and you can only sum modules over the same ground ring. – Verdruss Nov 02 '17 at 13:52
  • @DietrichBurde Does $a$ need to be an integer? I was under the impression that X only needs to be some abelian group (X,+), so $a[1]$ is just $a[1]$. Not true? – r1697464 Nov 02 '17 at 13:54
  • It's unclear to me how you could define scalar multiplication in a meaningful way to give this construction a module structure. I believe you've left some ambiguity in the base ring which needs to be clearly defined. – CyclotomicField Nov 02 '17 at 13:55
  • @Verdruss Yeah, that is what threw me as well. OK, so if not a module, then what object could I consider $M$ as? – r1697464 Nov 02 '17 at 13:55
  • @r1697464 Looks sort of like the cartesian product of two rings with $a$ and $b$ being placeholders for the first and second coordinates similar to the $i,j,k$ notation for vectors in $\mathbb{R}^3$ – CyclotomicField Nov 02 '17 at 13:56
  • @CyclotomicField Hm, true. If I consider it as a cartesian product, can I talk about a quotient with $N$? Would $M/N$ be something like $a(\mathbb{Z}_3,\mathbb{Z}_5)$? – r1697464 Nov 02 '17 at 13:59
  • (Thanks, you guys. I am new so sorry if my questions are stupid.) – r1697464 Nov 02 '17 at 14:00
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    There is a lot to address here even before you throw the question about the quotient into the mix. As it stands, it's not clear what the $N$ should mean here. – rschwieb Nov 02 '17 at 14:01
  • @rschwieb I see. So I guess, my first question is when would $M$ make sense? What if it is not $\mathbb{Z}_3$ and $\mathbb{Z}_5$, but other rings $R_1$ and $R_2$? What if $R_1 = \mathbb{R}[x]$ and $R_2 = \mathbb{R}[y]$? What if $N = (a-b).1$ in that case? Would that make sense? – r1697464 Nov 02 '17 at 14:06
  • @r1697464 These are pretty normal questions for someone unaccustomed to the strictness of mathematical definitions. Many problems are assigned as definition checks just to see if the student understands what the thing actually is. I would encourage you to go through the definition criteria methodically with a few examples like this one even if it seems trivial. The practice will be invaluable. – CyclotomicField Nov 02 '17 at 14:09
  • @CyclotomicField Thank you. I thought about going through the definitions, but I got stuck at the first stage since they only every talk about the one ring. So I wondered, what if there are two rings? If not a module, what is that? In particular, for my example to rschwieb above, with $R_1 = \mathbb{R}[x]$ and $R_2 = \mathbb{R}[y]$, it seems intuitively that $M = af(x) + bg(y)$, and $N=(b-a)$, and $M/N = a(f(x) + g(y))$ should all make sense. Anyway, I do appreciate the thoughts everyone is throwing out, thanks. – r1697464 Nov 02 '17 at 14:15
  • @r1697464 Not to stand in the way of imagination, but if you are getting hung up on a "what if" your material doesn't even discuss, and that prevents you from having a chance to see the rest of the basic material, I would question the wisdom of lingering too long or burrowing too deep. Simply asking "what if I have two modules over different rings" or something like that is a big enough question already. – rschwieb Nov 02 '17 at 14:39
  • @r1697464 I don't really like to address questions like "I have this random idea I thought of... when would it make sense?" unless they really have a kernel of direction in them. I've tried to find a little bit of direction to offer you in the solution I gave. – rschwieb Nov 02 '17 at 14:41
  • @rschwieb Sure, thanks for your time. I'll keep the question open just in case someone else strolls by and has any other thoughts. – r1697464 Nov 02 '17 at 14:43

1 Answers1

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For the first $+$ to make sense, you would have to be adding them in some category, and one that makes sense immediately in this case is the category of abelian groups (or $\mathbb Z$-modules, if you prefer.) The action of $\mathbb Z$ on any particular element is just "repeated addition."

But this does not really allow $a,b$ to be "a basis" since $(a\cdot 1_{\mathbb Z_3})3=0$, for example. But what you're describing sounds an awful lot like the abelian group $\mathbb Z_3\times\mathbb Z_5$ otherwise.

Obviously, what you're describing isn't a $\mathbb Z_3$ or $\mathbb Z_5$ module (for example, $(a\cdot 1_{\mathbb Z_3})5=a\cdot 2\cdot 1_{\mathbb Z_3}$ which ought to be zero if it is a $\mathbb Z_5$ module, but isn't.)

You can however, make it into a $\mathbb Z_{15}$ module, but the "basis" still doesn't work since, for example, $(a\cdot 1_{\mathbb Z_3})3=0$ but $3\neq 0$ in $\mathbb Z_{15}$.

rschwieb
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  • I had to read your answer slowly. So, alternatively, what if I consider the example in the comments above? What if instead I consider infinite rings $\mathbb{R}[x]$ and $\mathbb{R}[y]$? Then, $a$ and $b$ is a basis if I am not wrong. Then does the setup make sense? Thank you. – r1697464 Nov 02 '17 at 14:40
  • @r1697464 You can't just mash any two modules together and expect to get anything nicer than an abelian group, no. And the only abelian groups that are going to have to have bases are free abelian groups (ones that are the direct sum of copies of $\mathbb Z$. – rschwieb Nov 02 '17 at 14:44
  • @rschweib Thanks. By the way, I just found this: https://math.stackexchange.com/questions/2142672/modules-over-direct-sum-of-different-rings They say that I could view $M_1 \oplus M_2$ as a module over $R_1 \times R_2$, with $M_1 = aR_1$ and $M_2 = aR_2$. The scalar multiplication they give seems to satisfy the requirements in the definition of modules. – r1697464 Nov 02 '17 at 15:01
  • @r1697464 Yes, you could go that way too. – rschwieb Nov 02 '17 at 16:24
  • Thank you for your time. – r1697464 Nov 02 '17 at 17:13