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Show that $\mathbb{Z}[\sqrt{2}]$ is a commutative ring

Do I just need to prove that it commutes for addition and multiplication?

Also, how would I find the units of this ring?

Eugene Zhang
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Gary
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  • Hard to know how much to prove. Probably closure under addition and multiplication, and depending on what definition of ring one uses, existence of at least one unit. All the rest is inherited from $\mathbb{R}$. As to the units, you may be expected to recall, not prove that you have described the full list. – André Nicolas Nov 02 '15 at 22:28
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    What is your definition of $\mathbb{Z}[\sqrt{2}]$? There are several different (equivalent) definitions. – Eric Wofsey Nov 03 '15 at 01:22

2 Answers2

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You just have to show that it is closed under addition and multiplication, then it is automatically a subring of the real numbers, which is already a commutative ring. But this is straightforward.
The units can be found by applying the Euclidean norm function to the equation $xy=1$, see the link given above.

Dietrich Burde
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$\mathbb{Z}[\sqrt{2}]$ is a commutative ring because it is the image of the ring homomorphism $\mathbb{Z}[x] \to \mathbb R$ that takes $x$ to $\sqrt2$.

The choice of $\sqrt2$ does not play a key role here: any real number will do. However, the choice of $\sqrt2$ does play a key role when it comes to finding units.

lhf
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