- #1

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Is Z[sqrt2] defined as {a+b(sqrt2)|a,b in Z}?

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- Thread starter pivoxa15
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- #1

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Is Z[sqrt2] defined as {a+b(sqrt2)|a,b in Z}?

- #2

HallsofIvy

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Your question really is- can that be expressed as {a+ b(sqrt(2))| a,b in Z}.

To answer that question, you have to see if it is a ring. Certainly, since rings are closed under multiplication, it must contain b(sqrt(2)) for any integer b. Since rings are closed under addition, it must also contain any number of the form a+ b(sqrt(2)). Is that enough? Is this closed under addition? Is it closed under multiplication? Since the commutative, associative and distributive laws are true for all real numbers and this set is clearly a subset of the real numbers, you don't need to show those separately.

- #3

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Yes {a+ b(sqrt(2))| a,b in Z} does satisfy being a ring so Z[sqrt2] can be expressed as that set.

- #4

HallsofIvy

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Good. What about Z[cuberoot(2)]? Is that equal to {a+ bcuberoot(2) |a, b in Z}?

- #5

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No because it is not closed under multiplication.

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