# Is that Ring a field?

Given a commutative Ring $R$ of ordered pairs $(x,y)$ of reals $x,y$ with addition and multiplication defined in the following way.

$$(x,y) + (u,v) = (x+u,y+v)$$ $$(x,y).(u,v) = (xu-yv,xv + yu)$$

I already showed that $R$ is an integral domain , now i need to show to prove $R$ is a field or not .

If $R$ is a field then every nonzero element $(x,y) \in R$ have a multiplicative inverse , which is $(x,y)(m,n) = (xm-yn,xn + ym) = (1,0)$

$(1,0)$ is the ring unity and $(m,n)$ is the multiplicative inverse and $xm - yn =1$, $xn + ym = 0$ how do we show that such $(m,n)$ exists or not ?

• Ordered pairs of what? – user98602 Nov 24 '14 at 4:36
• Where are those operations inside the ordered pairs happening? – Timbuc Nov 24 '14 at 4:38
• of reals , $x,y$ both belongs to the reals – alkabary Nov 24 '14 at 4:38
• Hint: $\mathbb Z[i] = \{a + ib\in\mathbb C| a,b \in \mathbb Z\}$ – Fra Nov 24 '14 at 4:38
• Ah, the reals! Yes, that thing is a field isomorphic with the complex $\;\Bbb C\;$ . For the inverse:$$(a,b)\neq (0,0)\implies (a,b)^{-1}=\left(\frac a{a^2+b^2}\;,\;\;-\frac b{a^2+b^2}\right)$$ – Timbuc Nov 24 '14 at 4:40

Alternatively, you can recognize that $(x,y)$ is kind of like $x + yi \in \mathbb{C}$ with the given operations.

From your definition, $$(u, v) + (x,y) = (u + x, v + y)$$ and $$(u, v)\cdot (x, y) = (ux - vy, uy + vx).$$

But this is the same behavior as $$(u + vi) + (x + yi) = (u + x) + (v + y)i$$ and $$(u + vi)(x + yi) = ux + uyi + vxi + vyi^{2} = ux + uyi + vxi - vy = (ux - vy) + (uy + vx)i$$

in the complex numbers.

But if we have a non-zero complex number $x + yi$, what is its multiplicative inverse? Well, the multiplicative inverse should be $\dfrac{1}{x + yi}$. But how do we express this as $u + vi$? Well, $\dfrac{1}{x + yi} = \dfrac{1}{x + yi} \cdot \dfrac{x - yi}{x - yi} = \dfrac{x - yi}{x^{2} + y^{2}} = \dfrac{x}{x^{2} + y^{2}} + \dfrac{-y}{x^{2} + y^{2}}i$.

Since the space you are dealing with "acts" like the complex numbers, that means the multiplicative inverse of $(x,y)$ should be $(\dfrac{x}{x^{2} + y^{2}}, \dfrac{-y}{x^{2} + y^{2}})$. You should check that this multiplied by $(x,y)$ gives you $1$.

Solve for $m$ and $n$ to get $n=\frac {-y}{x^2+y^2}$ and similarly for $m$ .Since $x,y$ are not both zero inverse exists