When are cancellations allowed in ring?

Question

During the lecture my professored mentioned something like "cancellation is perfectly fine in a ring when dealing with addition, but not with multiplication!". The example he gave was that, in $\mathbb{Z}_6$, $[3]\times[2]=[3]\times[0]=[0]$, but obviously $[2]\neq[0]$. I get that part.

But why cancellation in addition is valid? I don't quite get that. To give an example, when proving that $f(0)=0$ if $f$ is an isomorphism, we have something like $f(0)=f(0)+f(0) \implies f(0)=0$. Why can we just move one $f(0)$ to the other side?

I was suspecting that is because we can always add $-f(0)$ to both side of the equation and then that would let the left hand side equal to zero. But then the question becomes "why can we add the same thing on both side of the equation it still holds?"

Anyone can help me with this? Thanks!

Answer 1

Your suspicion is correct: you can add $-f(0)$ to both sides, and after applying some ring axioms, the $f(0)$ disappears. You can't do this for multiplication because there aren't necessarily inverses.

As for "why can we add/subtract/multiply/etc things to both sides of the equation", it's just a property of functions. For example, let's show that $a = b \implies a + c = b + c$. Define $f(x)$ to be $x + c$. Since $a = b$, we know that $f(a) = f(b)$, because functions are single-valued. So $a + c = b + c$.

You don't really need to define $f$ to do this, since $+$ is a well-defined function already, but doing so might make it clearer to you.

Answer 2

In a ring $(R,+,\times)$ we have $(R,+)$ is an abelian group where one of his axioms is: each element $a$ in this group has a symmetric element denoted by $-a$ such that $$a+(-a)=-a+a=0$$ hence

$$a+b=a+c\implies -a+a+b=-a+a+c\implies b=c$$

Answer 3

"why can we add the same thing on both side of the equation it still holds?"

Because addition is a function from $R\times R\to R$.

In particular the restriction to $\{x\}\times R\to R$ Is a function, and actually you can view this as a function from $R\to R$.

That is why adding an element preserves the sides of an equality: it is a functional relationship.

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More generally, cancellations are going to hold in any group for exactly the same reason. In a monoid, like the multiplicative monoid of a ring, you're always going to be able to cancel an element with an inverse, for the same reason.

Finally, one more tidbit about multiplicative cancellation. Even more generally, rings called "integral domains" admit multiplicative cancellation of nonzero elements even if they don't have inverses.

Answer 4

The reason you can add something to both sides of an equation and have it still be true is a special case of the "substitution property of equality", which is an axiom of equality as inherent as reflexivity, symmetry and transitivity. It says that if $b=c$, and $F(x)$ is any expression, then $F(b)=F(c)$. In this case, letting $F(x)=a+x$, this means $b=c$ implies $a+b=a+c$.