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

1. Apr 1, 2007

### pivoxa15

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

2. Apr 1, 2007

### HallsofIvy

Staff Emeritus
No, it's not defined that way. Since Z is a ring (but not a field), Z[sqrt(2)] is defined as "the smallest ring containing all of Z and sqrt(2)".

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. Apr 1, 2007

### pivoxa15

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

4. Apr 2, 2007

### HallsofIvy

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

5. Apr 2, 2007

### pivoxa15

No because it is not closed under multiplication.