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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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2 Answers 2

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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Yes, thank you. So how can we approach externally? I mean when I take any element in the intersection of all rings containing $\sqrt{2}$ and rational numbers, how can show that it is of the form a+b$\sqrt{2}$? –  suergin Nov 24 '12 at 16:20
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Because an element contained in all such rings is also contained in this special such ring. –  Hagen von Eitzen Nov 24 '12 at 16:39

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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