Proof, that $(\mathbb R×\mathbb R;+,*) $ is a ring, but not a field How do I prove, that $(\mathbb R×\mathbb R;+,*) $ is a ring, but not a field, where the $+$ and $*$   operations are: $(a,b)+(c,d):=(a+c,b+d)$ and $(a,b)*(c,d):=(ac,bd)$?
For the solution: so I would have to first  show, that $(\mathbb R×\mathbb R;+,*) $ is a ring, I have to prove that using the definition of a ring:


*

*$(\mathbb R×\mathbb R;+)$ has to be commutative group

*$(\mathbb R×\mathbb R;*)$ has to be a semigroup

*$*$ must be distributive over $+$ (from both sides)



If 1. 2. and 3. can be proven, then  $(\mathbb R×\mathbb R;+,*) $ is a ring.


*

*here I have to prove that  $(\mathbb R×\mathbb R;+) $ is an algebraic structure where $+$ is associative and commutative; the identity element is $0$ and that all elements have an inverse

*$*$ has to be associative

*? (How do I prove distributivity in this particular example?)



Now I have to prove that  $(\mathbb R×\mathbb R;+,*) $ is not a field. (How do I do that?)
Also how can I show whether or not  $(\mathbb R×\mathbb R;+,*) $ is a commutative ring?
 A: Hints: 
To show it's a ring, just write out what those definitions mean in longhand (i.e. (1) means $(a,b) + (c,d) = (c,d) + (a,b)$) and then demonstrate that they hold.
I guess maybe you should determine what the additive and multiplicative identities are; it is "obvious" that they are $(0,0)$ and $(1,1)$, but you can easily demonstrate that these are correct using the definitions of $+$ and $*$ and the fact that identities are unique.
If it's a ring, but not a field, that probably means that multiplication isn't always invertible. That is, if you find one $(a,b) \neq (0,0)$ such that there is no $(c,d)$ with $(a,b)*(c,d) = (1,1)$, then you have it.
To show it is commutative, just demonstrate that $(a,b)*(c,d) = (c,d)*(a,b)$; this is straightforward (it relies on the commutativity of multiplication in $\mathbb{R}$).
A: *

*In more general, for any rings $R,S$, their direct product $R\times S$ will be again a ring when the operations are defined coordinate-wise, like in this example. For the distributivity:
$$(r,s)\cdot\left((r_1,s_1)\,+\,(r_2,s_2)\right)\ =\ 
(r,s)\cdot\left(r_1+r_2,\,s_1+s_2\right)\ =\\
=\ \left(r\cdot(r_1+s_1),\,s\cdot(s_1+s_2)\right)$$
and now use distributicity in $R$ and $S$ (which now coincide with $\Bbb R$).

*There are zero divisors in $\Bbb R\times\Bbb R$, namely e.g. 
$$(1,0)\cdot(0,1)=(0,0)\,.$$

*It is commutative, as $(a,b)(c,d)=(ac,bd)=(ca,db)=(c,d)(a,b)$ holds.

