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I am recently learning about fields and frequently get stuck whole trying to prove some properties that drop out of the axioms. I'll give an example to try and see how too go about doing it.

Let $m,n \in \mathbb{F}$ some field. Then prove $(-m)n=-(mn)=-mn$. Now I know $(-m)$ is the unique element of the field such that $(-m)+m=0$ and $-mn$ is the unique element of the field such that $(-mn)+mn=0$

I have so far done $(-m)n=(-m1)n=(-1\cdot(mn))=-1\cdot (mn)$ but I don't know how to just write this is $=-(mn).$

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  • $\begingroup$ How did you get $(-m1)n=-1\cdot(mn)$? $\endgroup$
    – Eric Wofsey
    Mar 11, 2016 at 8:16
  • $\begingroup$ Associativity of multiplication in the ring. Is that step wrong? $\endgroup$
    – Gareth
    Mar 11, 2016 at 8:17
  • $\begingroup$ In general, when you want to proof $x=-y$, you always should look at $x+y$ and show that it is zero. In those basic consequences from the axioms, this is always the right choice to find a proof. $\endgroup$
    – MooS
    Mar 11, 2016 at 8:18
  • $\begingroup$ What specific elements are you applying associativity to? I don't see a "$-1$" anywhere on the left-hand side. (The step is correct, but it needs more justification, and the method of justification will probably be helpful for the last step too.) $\endgroup$
    – Eric Wofsey
    Mar 11, 2016 at 8:18
  • $\begingroup$ So we want to start with $(-m)n+(-mn)$ and try to show it's zero? $\endgroup$
    – Gareth
    Mar 11, 2016 at 8:19

2 Answers 2

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Hint: $$(−1)\cdot(mn) + mn = (−1)\cdot(mn) + 1\cdot mn = ((-1) +1)mn = \cdots$$

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From $m+(-m)=0$ ( definition of the opposite element), by distributivity we have: $$ (m+(-m))\times n=0 \times n \quad \iff \quad m\times n+(-m)\times n=0\times n $$

since $0 \times n=0 \quad \forall n$, this is:

$$ m\times n+(-m)\times n=0 $$

and, by definition and unicity of the opposite this means that $$ -(mn)=(-m)n $$

You can do the same for $m\times (-n)$.

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