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Consider the subset $\mathbb{Q}[\sqrt2]:=\{x+\sqrt{2}y\mid x,y\in\mathbb{Q}\}$ of $\mathbb{R}$, together with the addition and multiplication usual/common in $\mathbb{R}$

(Does this mean that the operations $(+)$ and $(\cdot)$ are from $\mathbb{R}$ ?).

Is $\mathbb{Q}$ a field? Prove your testimony.

If you want to find a neutral element $e$ for multiplication, such that $e\circ x=x$, you will get stuck:

\begin{align} e\circ x&=x\\ (x+y\sqrt2)\cdot(z+w\sqrt2)&=z+w\sqrt2\\ e=(x+y\sqrt2)&=1\qquad\qquad \forall x\neq0 \end{align}

If you want to find an inverse element for multiplication $a^{-1}$ such that $a\circ a^{-1}=e$, you will get stuck: \begin{align} (a+\sqrt2b)\cdot a^{-1}&=1\\ a^{-1}&=\frac{1}{a+b\sqrt2} \qquad\qquad\forall a\neq0 \end{align} Is this the point where I can say, that $\mathbb{Q}[\sqrt2]$ is not a field, because you can't find a neutral and inverse element for all elements in $\mathbb{Q}$?

Doesbaddel
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1 Answers1

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Actually, it is a field. A neutral element for the multiplicatio is $1\left(=1+0\sqrt2\right)$. And if $a+b\sqrt 2\neq0$, then$$\frac1{a+b\sqrt2}=\frac{a-b\sqrt2}{a^2-2b^2}=\frac a{a^2-2b^2}-\frac b{a^2-2b^2}\sqrt2\in\mathbb Q\left[\sqrt2\right].$$

José Carlos Santos
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