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In a $R$-module if $a,b\in R$ and $a\neq0, $$b \neq 0$ can it be that $ab=0$? All the modules I have seen do not allow this so I wonder if it is the case generally.

The only axiom that I see which could prevent this is (when $x$ is a vector):

$$(a\cdot_Rb)x=a\cdot_R(bx)$$

Which is odd, since any vector multiplied by elements would have to equal zero, since $0\cdot x=0$. I can't come up with a proof so perhaps someone here can help.

Dole
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    Well, what's $R$? If it is a general ring, then sure. Look at $R=\mathbb Z/6\mathbb Z$ and let $a=2,b=3$. – lulu Apr 12 '18 at 19:59
  • This is not a question about the module, it's a question about the ring $R$. – Ethan Bolker Apr 12 '18 at 20:00
  • The rings where $ab\ne0$ for all non-zero $a$ and $b$ are called (integral) domains. This definition is usually covered in the same chapter where the book covers the definition of ring. Endowing a set with a structure of $R$-module does not affect this fact. –  Apr 12 '18 at 20:00
  • Please note that you have caused some confusion by not stating your question clearly in the body of the question: you have given information in the title that is not in the body (namely the information that you are apparently only interested in products of scalars). – Rob Arthan Apr 12 '18 at 20:27

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You seem to have the ring and the module a bit mixed up in your question: are you interested in $R$ itself as an $R$-module or a more general $R$-module $M$ (say)? In the latter case, if you take $R = \Bbb{Z}$ and $M = \Bbb{Z}/2\Bbb{Z}$, then you have $2x = 0$ for any $x \in M$. If you want $M$ to be $R$ itself, then $R = \Bbb{Z}/m\Bbb{Z}$ provides another example for any non-prime non-unit $m$ (because if $m = ij$ is a non-trivial factorisation of $m$, the product $[i][j]$ of the equivalence classes of $i$ and $j$ will be the equivalence class $[m] = 0$ in $R$).

In a ring $R$, an element $a \in R$ such that there is a $b \in R \setminus \{0\}$ with $ab = 0$ is called a zero divisor. Many rings have zero divisors. Rings that do not have zero divisors are called integral domains.

In an $R$-module $M$, given an element $x$, the set of $a \in R$ such that $ax = 0$ comprise an ideal called the annihilator of $M$. Unless $R = 0$ or $R$ is a field, there will exist non-trivial $R$-modules with non-trivial annihilators. (Because if $I$ is an ideal of $R$ other than $0$ or $R$, $R/I$ is a non-trivial $R$-module with $I$ as its annihilator.)

Rob Arthan
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  • Thank you, do you think the module can be non-trivial, if the ring is commutative? – Dole Apr 12 '18 at 20:19
  • A non-trivial module (in the sense I meant above) is one with more than one element. Any (non-trivial!) ring $R$ has many non-trivial modules, e.g., $R^n$ for any positive integer $n$. – Rob Arthan Apr 12 '18 at 20:22
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I note that general elements of a module do not have a multiplication, so $a \cdot b$ need not make sense there.

What you have actually asked is whether two elements of the ring $R$ can have a product which is zero. Yes. In the ring $\mathbb{Z}/6\mathbb{Z}$, $2 \neq 0$, $3 \neq 0$, and $2 \cdot 3 \cong 6 \cong 0$. We say that $\mathbb{Z}/6\mathbb{Z}$ has zero divisors. We can treat $\mathbb{Z}/6\mathbb{Z}$ as a module over itself, yielding a module which does what you are describing.

Normally, the definition of module does not exclude rings with zero divisors. If we wished to exclude zero divisors, one way is to say "module over an integral domain". (This is a restriction of "module" = "module over a ring", in that an integral domain is a commutative ring and it has no zero divisors.)

Note that in any ring with zero divisor, $z$, and (at least) a unit, $u$, there is a $z'$ such that $z z' = 0$. But then $(z+u)z' = zz'+uz' = uz'$, so right cancellation doesn't hold. Be a little careful when there are zero divisors around.

Eric Towers
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