I have the following problem and I'm stuck in the second part:

Let $K$ be a field. We define in $K \times K$ the next operations: $$(a, b) + (c, d) := (a + c, b + d)$$ $$(a, b) · (c, d) := (ac − bd, ad + bc)$$

  1. Prove that $K\times K$ is a commutative and unitary ring with the given operations. Done
  2. Prove that $K \times K$ is a field iff $K$ satisfies that for all $a,b\in K$ that $a^2 + b^2 =0$ then $a=b=0$.

Any hints? Thanks!

  • 1
    $\begingroup$ Do you know what the multiplicative identity is? $\endgroup$ – user159517 Jan 15 '17 at 15:38
  • 2
    $\begingroup$ These are basically the operations you put on $\Bbb R^2$ to make it into $\Bbb C$. So, I guess you should try using the formula for the inverse in $\Bbb C$ and see what you get. $\endgroup$ – user228113 Jan 15 '17 at 15:39
  • $\begingroup$ Hint: Think about how you take inverses in the complex numbers and how that relates to the condition here $\endgroup$ – Tobias Kildetoft Jan 15 '17 at 15:39
  • $\begingroup$ Thanks for the answers, done! $\endgroup$ – 2pac Jan 15 '17 at 15:55
  • $\begingroup$ Are you familiar with quotient rings such $\,K[x]/f(x) = K[x]\bmod f(x)?\, $ If so then first show that your ring is isomorphic to $\,K[x]/(x^2+1)\,$ and all follows easily and intuittively, e.g.see here. If not, then you can instead use congruences to get the same effect - just as for integers $\endgroup$ – Bill Dubuque Jan 15 '17 at 16:25

Since $(a,b)\cdot(1,0)=(a,b)$, $(1,0)$ is the identity element of $K\times K$.

If $K\times K$ is field, suppose $(c,d)$ be the inverse of $(a,b)$. Then $$ (a,b)\cdot(c,d)=(ac-bd,ad+bc)=(1,0) $$ Hence we have $$ac-bd=1\tag1$$ $$ad+bc=0\tag2$$ So there are $acb-b^2d=b$ from $(1)$ and $bc=-ad$ from $(2)$. Plug latter into former we have $$ -a^2d-b^2d=-d(a^2+b^2)=b $$ So if $a^2+b^2=0$, then $b=0$. And $a^2=0$ means $a=0$.

Conversely suppose for all $a,b\in K$ that $a^2 + b^2 =0$ then $a=b=0$. Then for $(a,b)\neq 0 \:(a,b\in K)$, there is $a^2 + b^2 \neq 0$ for otherwise $(a,b)=(0,0)=0$. Thus $(a^2 + b^2)^{-1}$ exists in $K$. By solving $(1)$ and $(2)$, we have $$ c=a(a^2 + b^2)^{-1}, \quad d=-b(a^2 + b^2)^{-1} $$ Hence $(a,b)\cdot(c,d)=(1,0)$ and $(c,d)$ is the inverse of $(a,b)$, i.e. $K\times K$ is field.


Suppose $K\times K$ is a field. If $(a,b)\ne0$, then also $(a,-b)\ne0$ and so $$ (a,b)(a,-b)\ne0 $$ This proves one direction. For the converse, think to the complex numbers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.