Show that $A$ satisfies all the axioms to be a commutative ring with unity.

Question

Show that $A$ satisfies all the axioms to be a commutative ring with unity. Indicate the zero element, the unity, and the negative of an arbitrary a.

A is the set $ℤ$ of the integers, with the following "addition" $⊕$ and "multiplication" $⊙$ :

$$a \oplus b = a + b - 1$$ $$a ⊙ b = ab - (a+b) + 2$$

In regards to this question, I'm mostly struggling with how to apply these axioms with the given operations.

For example, to satisfy axiom 2, does the following approach make sense?

$Let\ a, b, c ∈ A$

$Then:$

$a = pq - (p + q) + 2$

$b = st - (s + t) + 2$

$c = xy - (x + y) + 2$

$And:$

$(a ⊙ b) ⊙ c = ([pq - (p + q) + 2] · [(s + t) + 2]) · [(x + y) + 2]$

$a ⊙ (b ⊙ c) = [(p + q) + 2] · ([st - (s + t) + 2] · [xy - (x + y) + 2])$

Answer 1

Not really.

if $a,b,c\in A$, then $a$, $b$, and $c$ are integers. That is all. You don’t express them as the result of products, which is what you are trying to do. They are just integers.

You need to show that the operations defined, $\oplus$ and $\odot$, satisfy the relevant axioms. For example, you need to show that $\oplus$ is associative; that is, that $$(a\oplus b)\oplus c = a\oplus (b\oplus c).$$ To do that, you need to evaluate each side and verify you get the same integer. Indeed, by definition, $$\begin{align*} (a\oplus b)\oplus c &= (a+b-1)\oplus c\\ &= (a+b-1)+c - 1\\ &= a+b+c-2. \end{align*}$$ where “$+$” and $-$ and $2$ are just the usual integer operations and numbers. On the other hand, $$\begin{align*} a\oplus(b\oplus c) &= a\oplus(b+c-1)\\ &= a+(b+c-1) - 1\\ &= a+b+c-2. \end{align*}$$ So you get the same result in both cases, hence $(a\oplus b)\oplus c = a\oplus(b\oplus c)$. So this new operation, $\oplus$, is associative.

Then you need to check the rest of the axioms: $\oplus$ is commutative, has an identity element (it need not be the usual $0$ form the integers; just some integer $z$ such that $a\oplus z=a$ for all integers $a$), additive inverses (again, it need not be $-a$; but given $a$, some $b$ such that $a\oplus b = z$). That $\odot$ is associative, distributes over $\oplus$ on both sides, and has an identity element (which need not be the usual $1$ from $\mathbb{Z}$, just some $u\in\mathbb{Z}$ such that $a\odot u = u\odot a = a$ for all integers $a$.