Let $\mathbb{R}^2$ be provided with component-wise multiplication $\cdot$ and component-wise addition $+$. Does $(\mathbb{R}^2,+,\cdot)$ form a field?
I would say no, because not all elements in $\mathbb{R}^2$ have an inverse element:
$\begin{align}x\circ x^{-1}&=e \qquad (x^{-1}:=\text{inverse element to }x)\\
(x,y)\cdot(x^{-1},y^{-1})&=(1,1)\\
(x^{-1},y^{-1})&=\left(\frac{1}{x},\frac{1}{y}\right)| \quad\text{Equation doesn't hold for } \mathbb{R}^2 \text{ but for } \mathbb{R}^2\setminus\{(0,0)\}
\end{align}$
No inverse exists for $(0,1)$ because $x^{-1}=(\frac10,\frac11)\notin\mathbb{R}^2 \implies \mathbb{R}^2$ is not a field.
Is this proof correct?
Best Answer
Yes, your proof by counterexample works.
Careful though, the equation does not hold for $\mathbb{R}^2\backslash{(0,0)}$, because, as you said, $(0,1)$ is not invertible.
For every non-zero element to be invertible, you would need to exclude every pair which has a zero in it. So $\mathbb{R}^2\backslash{(a,b)} \text{ such that } ab = 0$. (except the pair $(0,0)$).
Edit: as metamorphy pointed out though, that set is not closed under addition, so not a field.