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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:

  1. $(\mathbb R×\mathbb R;+)$ has to be commutative group
  2. $(\mathbb R×\mathbb R;*)$ has to be a semigroup
  3. $*$ must be distributive over $+$ (from both sides)

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

  1. 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
  2. $*$ has to be associative
  3. ? (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?

user26857
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user176791
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  • You can assume that $(\mathbb R; + ,)$ is a commutative ring, since the distributivity, commutativity etc of addition and multiplication are usually taken as axioms of arithmetic. To show it is not a field, it suffices to find $x, y\ne 0$ such that $xy = 0$ – Mathmo123 Sep 29 '14 at 19:25

2 Answers2

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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 $\mathbb{R}$).

BaronVT
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  1. 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 $\Bbb R$).
  2. There are zero divisors in $\Bbb R\times\Bbb R$, namely e.g. $$(1,0)\cdot(0,1)=(0,0)\,.$$
  3. It is commutative, as $(a,b)(c,d)=(ac,bd)=(ca,db)=(c,d)(a,b)$ holds.
Berci
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