Thread: Subrings

Which of the following are sets are subrings of the field R of real numbers?
a) A={m+n(2)^(1/2)|m,n in the integers and n is even}
b)B={m+n(2)^(1/2)|m,n in the integers and m is odd}
c)C={a+b(2)^(1/3)|a,b in the rationals}
d) D={a+b(3)^(1/3)+c(9)^(1/3)|a,b, c in the rationals}

My problem is getting started. I know a set R is a subring if it is closed under addition and multiplication, if a is in R, the -a is in R, R contains the identity

Just check the things you mentioned. For example, B is clearly not closed under addition. Also, it is not necessarily required that a subring contain 1, although it's possible that in your setting, rings are assumed to have multiplicative identities.

ok so m+n(2)^1/2+x+y(2)^1/2, m is even so call it 2b
2b+n(2)^1/2+2a+y(2)^1/2
2(a+b)+(n+y)2^1/2 Ok that makes sense for checking closure of addition. I think I can figure out multiplication. Now what about the a and -a stuff?
a is in m+n(2)^1/2. Now I get stuck on going further.

Well it looks like you're trying to show that is closed under addition, but if so then you have misread the set. We need even for . However, your method is more or less correct once you flip those back around. We have , and this preserves the conditions of . So is closed under addition. Also notice that if , then is the additive inverse, and this is also an element of . So is closed under additive inverses. Note also that . So you just need to show that is closed under multiplication.

Let and be members of . If you multiply them, do you get an element of ? If so, then is a subring. Otherwise it is not.

Ok I follow all that except for why 1 is in A. Is that because if we let m=1 and n=0?

Originally Posted by kathrynmath

Ok I follow all that except for why 1 is in A. Is that because if we let m=1 and n=0?

Yes, that is correct. is even and so .

D isn't a subring, right?

Why isn't a subring?

because it's not closed under multiplication. I got that it was closed under addition, but not closed under multiplication