Showing $\mathbb{Q} \times \mathbb{Q}$ is not a field

Question

I am revising and have come across the question

Show that $\mathbb{Q} \times \mathbb{Q}$ with element-wise addition and multiplication is not a field

I don't understand how to go about this, do i use the fact that all non-zero elements in a field are units and then try and obtain a contradiction?

Answer 1

You know that in a field $\;ab=0\iff a=0\;\;or\;\;b=0\;$ . Now try with $\;(1,0)\;,\;\;(0,1)\;$ in your case

Answer 2

$(1, 0) \ne (0, 0) \ne (0, 1)$, but $(1,0) \cdot (0,1) = \dots$

Answer 3

You can do it as you suggest. Consider say $(2,0)$ and note that after whatever multiplication the second coordinate will still be $0$.

Thus, you can never get $(1,1)$ the identity with respect to multiplication.

Note that the subset $\{(q,0) \colon q \in \mathbb{Q}\}$ would be a field, yet with a different identity element namely $(1,0)$. (This however is only an identity element relative to this subset and not the full set.)