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There's a Lemma concerning properties of rings I have to prove.

Prove the following properties of a ring $\langle R; +,-,0,\cdot,1\rangle$ and $a,b \in R$:

(i)$\hspace{2.98em}0a=a0=0$.
(ii)$\hspace{2.7em}(-a)b=-ab$.
(iii)$\hspace{2.5em}(-a)(-b)=ab$.
(iv)$\hspace{2.55em}$ If $R$ is non-trivial (i.e., if it has more than one element), then $1≠0$.

The exercises says that I'm allowed to use the statement (i) without proof.

My attempt for (ii):
Assumption 1 is in R:
$(-1)(-1) + (-1) = 0 = (1) + (-1)$ which is $(-1)^2 = (1)$? Is that even a proof?

I don't know about the other statements though.

Thanks I advance. I really appreciate it.

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  • $\begingroup$ By the way, (i) can be proved as well. Since $0 = 0+0$, $0a = (0+0)a = 0a + 0a$, which implies $0a = 0$. The other equality $a0 = 0$ can be proved similarly. $\endgroup$ – aras Nov 22 '15 at 13:54
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It is quite unclear to me what you are trying to do there, in particular it seems you want to use $(-1)(-1)=1$ which is a special case of what you should show in (iii). Moreover, there is no need to assume $1$ is in $R$, it appears your definition enforces this.

Now, to show (ii) it can be a good idea to get clear what is even written there, it seems so natural that it can be hard to see what the point is.

Note that $-ab$ is the additive inverse of $ab$, while $(-a)b$ is the product of $-a$, the additive inverse of $a$, and $b$.

You need to show that the additive inverse of $ab$ is $(-a)b$. Thus add $ab$ and $(-a)b$ and see if you get $0$. If this is true, then you have shown that $(-a)b$ is the additive inverse of $ab$, that is $-ab$. (This assumes you know already that an additive inverse is unique.)

To see this note $ab +(-a)b = (a+(-a))b = 0 b = 0$, where you used the distributive law and (i).

For (iii) you basically use (ii) twice (and the fact that $-(-c)=c$).

For (iv), take any $a$ in $R$. Then if $1 = 0$ then of course $1a = 0 a$. But the former is $a$ (by definition of $1$) while the latter is $0$ by (i). Thus you get $a= 0 $ for all $a$ in $R$, that is the ring is trivial.

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(ii) $(-a)b=-(ab) \Leftrightarrow (-a)b+ab=0 \Leftrightarrow ((-a)+a)b=0 \Leftrightarrow 0·b=0$ which is true by (i)

(iii) $(-a)(-b)=-a(-b)$ by (ii). But $-a(-b)=ab \Leftrightarrow ab+a(-b)=0 \Leftrightarrow a(b+(-b))=0 \Leftrightarrow a·0=0$ which is true by (i)

(iv) If $R$ is non trivial $\exists a\in R-\{0\}$. Then $(1·a)-a=0$. If $1=0$ then $1·a=0$ and hence $-a=0 \Leftrightarrow a=0$ which is not true.

We used the definition of $-a$, which is:

$-a$ is an element of $R$ such that $a+(-a)=0$

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Hint for (ii): show that $ab$ is the additive inverse of the RHS and the LHS. By uniqueness of inverses (you probably will have seen the proof of this), RHS = LHS.

Hint for (iii): Use (ii).

Hint for (iv): Prove the contrapositive. Let $R$ be a ring where $0=1$ and pick a $x \in R$. You should use the fact that $0=1$ to show that $R$ is trivial.

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