1. ## 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

2. 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.

3. 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.

4. Well it looks like you're trying to show that $A$ is closed under addition, but if so then you have misread the set. We need $n$ even for $m+n\sqrt{2}$. However, your method is more or less correct once you flip those back around. We have $(m_1+n_1\sqrt{2})+(m_2+n_2\sqrt{2})=(m_1+m_2)+(n_1 +n_2)\sqrt{2}$, and this preserves the conditions of $A$. So $A$ is closed under addition. Also notice that if $m+n\sqrt{2}\in A$, then $-m-n\sqrt{2}$ is the additive inverse, and this is also an element of $A$. So $A$ is closed under additive inverses. Note also that $1\in A$. So you just need to show that $A$ is closed under multiplication.

Let $m_1+n_1\sqrt{2}$ and $m_2+n_2\sqrt{2}$ be members of $A$. If you multiply them, do you get an element of $A$? If so, then $A$ is a subring. Otherwise it is not.

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

6. 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. $n=0$ is even and so $1\in A$.

7. D isn't a subring, right?

8. Why isn't $D$ a subring?

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