Is $\mathbb{R}^2$ with product defined like $(x, y)\cdot(a, b)=(ax, by)$, and usual addition, a field?

Question

It satisfies the definition of field. The multiplicative identity can be $(1,1)$. So, is it a field this way too (other than the obvious complex number way)?

Answer 1

Well! clearly not because by definition every non-zero element should be invertible but any element of the forms $(x,0)$ or $(0,y)$ are not invertible clearly. I hope this works.

Answer 2

$(1,0) \cdot (0,1)=(0,0)$ so you have zero divisors. It’s not even an integral domain.