2
$\begingroup$

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 ?

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

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 $1$.

$\endgroup$
6
$\begingroup$

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

$\endgroup$

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.