Quadratic Field Extension $\mathbb Q(\sqrt3)$

Detailed Solution

We are given $$ R=\{a+b\sqrt3:a,b\in\mathbb Q\}. $$ We need to determine whether $R$ is a field under the usual addition and multiplication. Step 1: Verify closure under addition. Let $$ x=a+b\sqrt3,\qquad y=c+d\sqrt3, $$ where $a,b,c,d\in\mathbb Q$. Then $$ x+y=(a+c)+(b+d)\sqrt3. $$ Since $a+c,b+d\in\mathbb Q$, we have $x+y\in R$. Step 2: Verify closure under multiplication. We obtain $$ \begin{aligned} xy&=(a+b\sqrt3)(c+d\sqrt3)\\ &=ac+3bd+(ad+bc)\sqrt3. \end{aligned} $$ Since $ac+3bd$ and $ad+bc$ are rational numbers, $xy\in R$. Step 3: Identify the additive and multiplicative identities. The additive identity is $$ 0=0+0\sqrt3\in R. $$ The multiplicative identity is $$ 1=1+0\sqrt3\in R. $$ Also, every element $a+b\sqrt3$ has an additive inverse $$ -(a+b\sqrt3)=-a-b\sqrt3\in R. $$ Step 4: Prove that every nonzero element has a multiplicative inverse. Let $$ x=a+b\sqrt3\neq0. $$ Multiplying by the conjugate, we get $$ \begin{aligned} (a+b\sqrt3)(a-b\sqrt3) &=a^2-3b^2. \end{aligned} $$ We claim that $$ a^2-3b^2\neq0. $$ Indeed, if $a^2-3b^2=0$, then $$ a^2=3b^2. $$ If $b\neq0$, this implies $$ \frac ab=\pm\sqrt3, $$ which is impossible because $a/b\in\mathbb Q$ and $\sqrt3$ is irrational. If $b=0$, then $a=0$, contradicting $x\neq0$. Therefore, $$ a^2-3b^2\neq0. $$ Now, $$ \begin{aligned} x^{-1} &=\frac{1}{a+b\sqrt3}\\ &=\frac{a-b\sqrt3}{a^2-3b^2}\\ &=\frac{a}{a^2-3b^2} -\frac{b}{a^2-3b^2}\sqrt3. \end{aligned} $$ Both coefficients are rational. Consequently, $$ x^{-1}\in R. $$ Thus every nonzero element of $R$ has a multiplicative inverse. Therefore, $$ \boxed{R=\mathbb Q(\sqrt3)\text{ is a field}.} $$ Hence option A is correct. Why the other options are incorrect: Option B: False. We have explicitly constructed the multiplicative inverse of every nonzero element of $R$. Option C: False. Restricting $a,b$ to integers gives $\mathbb Z[\sqrt3]$, which is not a field. For example, $2\in\mathbb Z[\sqrt3]$ is nonzero, but its inverse $1/2$ does not belong to $\mathbb Z[\sqrt3]$. Option D: False. The map $$ a+b\sqrt3\mapsto(a,b) $$ is not a ring isomorphism to $\mathbb Q\times\mathbb Q$ with componentwise multiplication. In particular, $R$ is a field, whereas $\mathbb Q\times\mathbb Q$ is not a field because $$ (1,0)(0,1)=(0,0) $$ with both factors nonzero. Thus the correct answer is $$ \boxed{\mathrm{A}}. $$ Important result: For any nonsquare integer $d$, the set $$ \mathbb Q(\sqrt d)= \{a+b\sqrt d:a,b\in\mathbb Q\} $$ is a quadratic field extension of $\mathbb Q$.