Prove the $(\mathbb {N},\oplus,\otimes)$ is ring

Question

On the set of natural numbers$\mathbb { N} $, define the operations $a \oplus b := \max(a,b)$ and $a\otimes b := a+b$ Is $(\mathbb {N},\oplus,\otimes)$ is ring? commutative ring with unity? Field?

My solution :

1- $(\mathbb {N},\oplus) $ is abelian groub because :

a. It is comutative $a \oplus b = \max(a,b)= \max(b,a) =b \oplus a$

b. It is associative $(a \oplus b) \oplus c =a \oplus( b\oplus c) =\max(a,b,c)$

C. The identity of element $a$is $a $ $$a \oplus a= \max(a,a)= a $$ d. The inverse also a $a \oplus a = \max(a,a)=a $

$(\mathbb {N},\times) $

1.it is comutative : $a\otimes b = a+b= b+a= b\otimes a$

2. $\otimes $has a identity and it is 0 $a\otimes 0 = a+0=a$

3. $\otimes $ has inverse and inverse of a is -a $a\otimes -a = a-a=0$

Distributive law: $a\otimes (b\oplus c) =a\otimes (max (b,c))= a+max( b,c) $

It is true ? If not, why?

Thanks

Answer 1

The identity for $(\mathbb{N}, \oplus)$ should be $e=0$.

Answer 2

The identity can't be dependent on $a$. Anyway you want to have $a = a \oplus e = \max(a,e)$. So for each $a$ we must have $a \ge e$, which in natural numbers is $0$ (or $1$ if you consider $\mathbb{N}$ as the set of positive integers). But then there aren't any inverses, as for $a \not = 0$ we have that $a \oplus b = \max(a,b) \ge a > 0$, so there isn't equality. Therefore $(\mathbb{N},\oplus)$ isn't a group.

Answer 3

As soon as you saw $a\oplus a=a$ for every $a$, you should have immediately seen that if this were a ring, $a=a\oplus a\oplus -a=a\oplus -a=0$, a contradiction if there exists $a\neq 0$. So it clearly can't be a ring.

But it is easy to see that it is a sub-semiring of the max-plus algebra.