# $K \times K$ is a field iff $K$ satisfies that for all $a,b\in K$ that $a^2 + b^2 =0$ then $a=b=0$.

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!

• Do you know what the multiplicative identity is? – user159517 Jan 15 '17 at 15:38
• 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. – user228113 Jan 15 '17 at 15:39
• Hint: Think about how you take inverses in the complex numbers and how that relates to the condition here – Tobias Kildetoft Jan 15 '17 at 15:39
• Thanks for the answers, done! – 2pac Jan 15 '17 at 15:55
• 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 – 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.