# Verifying a subring of R?

How do I verify that $\mathbb{Z}[\sqrt{2}] = \{ a +b\sqrt{2} \, | \, a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{R}$ ? I'm thinking that i have to show that it's a subgroup which is closed under multiplication, is that correct?

Also, I have to show that $\mathbb{Z}[\sqrt{2}]^*$ is infinite. It furthermore says I'm supposed to consider powers of $1+\sqrt{2}$ . I am, however, completely lost on this one. I thought $\mathbb{Z}^* = \{ -1,1 \}$, and therefore $\mathbb{Z}[\sqrt{2}]^*$ would be $={-\sqrt{2},-1,1,\sqrt{2}}$, but I'm obviously wrong.

• Cathrine, the reason that your formulas were not showing correctly was that you needed to enclose them between $symbols. If you click on the edit history you'll see the changes I made. Also, if I'm not mistaken, the "algebra" tag is no longer used. Instead you should use the "abstract algebra" tag for this kind of topics. – Adrián Barquero Nov 12 '12 at 2:39 • Oh, I'll remember that. Thank you so much for your help! – MBrown Nov 12 '12 at 2:42 ## 2 Answers I'm going to try to answer your second question, since showing that it's a subring only needs you to verify the conditions in the definition of a subring. It seems that you're having some trouble understanding what$\mathbb{Z}[\sqrt{2}]^*$is. By definition$\mathbb{Z}[\sqrt{2}]^*$is the group of units of the ring$\mathbb{Z}[\sqrt{2}]$. It consists of elements$\alpha = a + b \sqrt{2}$with$a, b \in \mathbb{Z}$such that$\alpha$has a multiplicative inverse in$\mathbb{Z}[\sqrt{2}]$. This just means that if$\alpha = a + b \sqrt{2}$is in$\mathbb{Z}[\sqrt{2}]^*$, then there must exist another element$\beta = c + d\sqrt{2}$in$\mathbb{Z}[\sqrt{2}]$such that$\alpha \beta = 1$. Another fact that might be useful is that since$\mathbb{Z}[\sqrt{2}]^*$is a group, then given any two elements$\alpha_1, \alpha_2 \in \mathbb{Z}[\sqrt{2}]^*$, then their product$\alpha_1 \alpha_2 \in \mathbb{Z}[\sqrt{2}]^*$also. Certainly$\pm 1 \in \mathbb{Z}[\sqrt{2}]^*$, but$\pm \sqrt{2} \notin \mathbb{Z}[\sqrt{2}]^*$because for$\pm \sqrt{2}$to be in$\mathbb{Z}[\sqrt{2}]^*$, there must be an element$c + d\sqrt{2}$with$c, d \in \mathbb{Z}$such that$\pm \sqrt{2}(c + d\sqrt{2}) = 1. But observe that \begin{align} \pm \sqrt{2}(c + d\sqrt{2}) = 1 &\iff \pm c\sqrt{2} + \pm 2d = 1 \\ &\iff c = 0 \quad d = \pm\frac{1}{2} \end{align} But then sinced = \pm \frac{1}{2} \notin \mathbb{Z}$, this shows that$\pm \sqrt{2} \notin \mathbb{Z}[\sqrt{2}]^*$. On the other hand, since$(1 + \sqrt{2})(-1 + \sqrt{2}) = 1$, then$1 + \sqrt{2} \in \mathbb{Z}[\sqrt{2}]^*$. Now, you have to show that$\mathbb{Z}[\sqrt{2}]^*$has infinitely many elements, and for this you're provided with the hint to consider powers of$1 + \sqrt{2}$. Well, since we already know that$1 + \sqrt{2} \in \mathbb{Z}[\sqrt{2}]^*$and that products of elements in$\mathbb{Z}[\sqrt{2}]^*$also lie in$\mathbb{Z}[\sqrt{2}]^*$then we can conclude that any power$(1 + \sqrt{2})^n \in \mathbb{Z}[\sqrt{2}]^*$. Finally, since$1 + \sqrt{2} > 1$then each succesive positive power gets bigger, so that we have actually showed infinitely many different elements in$\mathbb{Z}[\sqrt{2}]^*$$$1 < 1 + \sqrt{2} < (1 + \sqrt{2})^2 < (1 + \sqrt{2})^3 < \cdots$$ Given a ring$R$, a subset$S\subseteq R$is a subring if it contains the multiplicative identity of R and is closed under the "subtraction" and multiplication of$R$. For the second question, recall that$\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\,|\,a,b\in\mathbb{Z}\}$. So$\mathbb{Z}[\sqrt{2}]^*$is the set of elements of the form$a+b\sqrt{2}$($a,b\in\mathbb{Z}$) that have multiplicative inverses. So think about the hint: does$x=1+\sqrt{2}$have a multiplicative inverse? What about powers of$x\$?

• I understand now. Thank you very much! – MBrown Nov 12 '12 at 3:05
• Very glad to help – Bey Nov 12 '12 at 3:14