Can anyone prove why the smallest ring containing $\sqrt{2}$ and rational numbers is comprised of all the numbers of the form $a+b\sqrt{2}$ (with $a,b$ rational)?
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That ring must surely contain all numbers of the form $a+b\sqrt 2$ with $a,b\in\mathbb Q$ because these can be obtained by ring operations. Since that set is closed under addition and multiplication (because $(a+b\sqrt 2)+(c+d\sqrt2)=(a+c)+(b+d)\sqrt 2$ and $(a+b\sqrt2)\cdot(c+d\sqrt 2)=(ac+2bd)+(ad+bc)\sqrt 2$), it is already a ring, hence nothing bigger is needed. |
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Clearly, $Q[\sqrt2]$ is a ring that contains $\sqrt 2$ and all the rationals. Morevoer, every element of $Q[\sqrt2]$ belongs to that smallest ring that contains $\sqrt 2$ and all the rationals. (This is because if a,b are any two rational numbers, then $b\sqrt 2$ belongs to that ring, thus $a+b\sqrt 2$ belongs to that smallest ring) |
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