Is my solution correct on this Groups/Rings

Question

So basically, I'm still learning about Groups/Rings and I was wondering if my solution for this exercise is correct. Also how can I solve the b part of the exercise (since i'm not sure).

Exercise: On a set R, we have a defined $*$ operation as follows: for any $x,y \in R$ , we have $x*y = ax+by+c$, for some $a,b,c \in R$.

a)Find a,b,c such that $(R,*)$ is an abelian group

b)Find $c$ such that $(R, *, \times)$ is a ring, where '$\times$' is normal multiplication over R.

My Solution:

a) To prove that it's abelian group. I created the following equation $$x*y = y*x \implies ax+by+c = ay+bx+c \implies ax+by = ay+bx \implies ax - ay = bx - by \implies a(x-y) = b(x-y) \implies a = b$$ Therefore in order for $(R, *)$ to be an abelian group $a$ must be equal to $b$. Is this correct?

Also for the b part this is my thinking (haven't really solved it)

b) We have already proved that it's an abelian group if $a = b$ therefore now we need to prove closure with '$\times$', association with '$\times$' and distributive with '$\times$'. Is my thinking correct if yes how would I finish this solution or is there a different approach to this?

Thank you very much!

Answer 1

You are correct, that for * to be abelian $a= b$

But in order for it to be a group, the operation must also be associative, there must be an identity, and every element must have an inverse element.

I don't think associativity will be a problem, nonetheless you should verify.

But the identity...

There extists and $e$ such that ex = xe = x for all $x\in R$

$x*e = ax + ae + c = x\\ y*e = ay + ae + c = y\\ a(x-y) = x-y$

$a = 1$

$e = -c$

$x * x^{-1} = x + x^{-1} + c = e = -c\\ x^{-1} = -2c - x$

If $R$ is a ring, then multiplication distributes over addition. We will treat multiplication as ordinary multiplication.

$x(y*z) = xy * xz\\ x(y + z + c) = xy + xz + c\\ xy + xz + xc = xy + xz + c\\ c = 0$

Answer 2

I do not have enough reputation to comment, so I will post as an answer.

In order to determine that a pair $(\mathbb R, *)$ is an abelian group, we must first and foremost determine that it is in fact a group, then we can verify that it is abelian. To verify that it is a group we must have

• $\forall x,y\in\mathbb R: x*y\in\mathbb R$
• $\forall x,y,z\in\mathbb R: (x*y)*z=x*(y*z) $
• $\exists e\in\mathbb R\ \forall x \in\mathbb R : e*x = x*e$
• $\forall x\in\mathbb R \ \exists x^{-1}: x*x^{-1} = x^{-1}*x = e$

That is, we need to assure closure and the product should be associative. There should exist a neutral, or unit, element $e$, and every element should have an inverse element.

You have already made progress in solving part a), just make sure you take into account the above equations when determining the constants $a,b,c$!

As for part b), we must check that we fulfill the definition of a ring, that is $(\mathbb R,*,\times)$ should be such that $(\mathbb R,*)$ is an abelian group and moreover

• $\forall x,y,z\in \mathbb R: (x\times y)\times z = x\times(y\times z)$
• $\forall x,y,z \in \mathbb R:x\times(y*z) = (x\times z)*(x\times y)$
• $\exists u\in\mathbb R \ \forall x\in\mathbb R: u\times x = x\times u$

That is, in addition to require that we have an abelian group with respect to the operation $*$ we require that the operation $\times$ is associative and distributes over $*$. This can be done by solving the associated equations above, just as you already did in your attempt! A good strategy is always to state the definition (or axioms) and check them one by one.