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?

  • $\begingroup$ you have to show that $(a,b) \times (c,d) = (ac,bd)$ for any $(a,b),(c,d) \in \mathbb{Q} \times \mathbb{Q}$ is not the operation of a group $\endgroup$ – reuns Apr 24 '16 at 11:35
  • $\begingroup$ @user1952009 that's not true, you need to impose non-zero. $\endgroup$ – quid Apr 24 '16 at 11:39
  • 1
    $\begingroup$ @quid I understood, you meant on $\mathbb{Q} \times \mathbb{Q} \setminus \{(0,0)\}$, yes $\endgroup$ – reuns Apr 24 '16 at 11:48

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


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

  • $\begingroup$ what does this mean? $\endgroup$ – Robert Thompson Apr 24 '16 at 11:37
  • $\begingroup$ @RobertThompson it's a recommendantion to show that the structure contains nontrivial zero-divisors. $\endgroup$ – quid Apr 24 '16 at 11:38
  • $\begingroup$ You should have seen that in a field, a product is zero iff one of the factors is zero. Because if $a b = 0$, and $b \ne 0$, you can multiply by $b^{-1}$ to get $a = 0$. $\endgroup$ – Andreas Caranti Apr 24 '16 at 11:39

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.)


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