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If $R$ is a ring and $a, b, c$ are elements of $R$, then does $b=c$ always imply $ab=ac$, or do I need to prove it using ring axioms?

Bill Dubuque
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    This is always true because your multiplication is a well defined map. – Randall Mar 04 '19 at 20:02
  • You can prove it. Whether or not you need to prove it depends on what was requested (in the homework or exam). – Dietrich Burde Mar 04 '19 at 20:02
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    If $,f(x,y),$ is any function from $R^{\large 2}\to R$ then $,b = c,\Rightarrow, f(a,b) = f(a,c).,$ OP is the special case when $f$ is the ring multiplication operation. – Bill Dubuque Mar 04 '19 at 20:04
  • I agree with @DietrichBurde however unless this is the very first weeks of a very basic course it's unlikely you need to prove such a thing. But you should know why it is true. – Yanko Mar 04 '19 at 20:06
  • To clarify, I am not asked to prove this, but I merely need to use this to prove another result (namely that, if R is a ring with identity, and ''0'' = ''1'', then R has only 1 element). Thing is, it is my first time doing rings, so I am super careful with what I consider a ''known result'' and I think I must prove everything I use, even if it is basic algebra. –  Mar 04 '19 at 20:17
  • @DietrichBurde there was no typo, the implication i ask about is indeed b = c => ab = ac –  Mar 04 '19 at 20:18
  • In this case have a look at the (super careful) answer by Alan on your question about the trivial ring here. – Dietrich Burde Mar 04 '19 at 20:22

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Very, very late answer, but still hopefully useful for future users:

You might be able to prove this using the ring axioms (depending on how you interpret them), but the simplest way to think about it is to use the logical axioms ring theory is built upon.The principle that if $a=b$, then $f(a) = f(b)$ for any function $f$ is called the "substitution property of equality," and is an axiom of equality in first-order logic. (See this wikipedia article). Essentially, this encodes the idea that you can't differentiate between equal quantities, so functions had better agree on them. Here, if we take $f$ to be the map sending $x$ to $ax$, the result follows.

Brandon Harad
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  • As mentioned in the comments above, $ab=ac$ follows a fortiori from the definition of 'binary operation,' which most people would agree are part of the ring axioms. This would seem to contradict the claim that "you definitely can't prove this using the ring axioms." Maybe some hairs are being split here on what "being part of the axioms" is. – rschwieb Mar 20 '25 at 18:28
  • Good point. I have edited. – Brandon Harad Mar 21 '25 at 00:05
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. $\ \ $ – Bill Dubuque Mar 21 '25 at 00:42