# Determining whether a set is a ring?

Our homework problem asked us to determine whether the following set is a ring or not.

My friend is telling me that it's not a ring, but I'm a bit confused about which condition it doesn't satisfy?

b) closed under multiplication
c) associative under multiplication
d) Distributive property?

$$R = \{a+b∛2 | a,b ∈ \mathbb{Q}\}$$ under usual Addition and Multiplication?

Any help is appreciated! :)

• @Gage: why? I'm pretty sure $R$ is not closed under scalar multiplication. Just take $\sqrt[3]{2} \in R$. $\sqrt[3]{2}^{2}$ is not in $R$, unless I'm mistaken. Nov 14, 2014 at 2:26
• @AWertheim Good call it looked like $\mathbb{Q}[\sqrt[3]{2}]$ to me but I totally missed that
– user171177
Nov 14, 2014 at 2:29
• @Gage: a mistake I've made myself many times. Cheers :) Nov 14, 2014 at 2:30

The question "is this set a ring?" doesn't make sense. In the definition of a ring, you will see that a ring is not just a set - it comes equipped with operations $+,-,*$ and constants $0,1$ (satisfying the ring axioms). These operations always have to be specified. There is a fundamental difference between a ring $(R,+,-,*,0,1)$ and its underlying set $R$ (which is often forgotten because usually they are both denoted by the same symbol).
In this case, probably one should consider the operations induced from the ring $\mathbb{R}$ with the usual operations. In other words, we ask if $R = \{a + b \sqrt[3]{2} : a,b \in \mathbb{Q}\}$ is a subring of $\mathbb{R}$. As was already remarked in the comments, although it is closed under $+$, it is not closed under $*$. The subring generated by $\sqrt[3]{2}$ and $\mathbb{Q}$ is given by $\mathbb{Q}(\sqrt[3]{2}) = \{a + b \sqrt[3]{2} + c \sqrt[3]{2}^2 : a,b,c \in \mathbb{Q}\}$.