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

  • Thread starter pivoxa15
  • Start date
  • #1
2,255
1
Is Z[sqrt2] defined as {a+b(sqrt2)|a,b in Z}?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
963
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
2,255
1
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
Science Advisor
Homework Helper
41,833
963
Good. What about Z[cuberoot(2)]? Is that equal to {a+ bcuberoot(2) |a, b in Z}?
 
  • #5
2,255
1
No because it is not closed under multiplication.
 

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

Replies
1
Views
3K
Replies
3
Views
1K
  • Last Post
Replies
0
Views
674
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
11
Views
5K
Replies
7
Views
10K
Top