Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.

Question

Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.

My professor has gone over how to prove a subring, but I am not sure how to prove a ring.

This is what I have so far, but I am not sure if it is similar to proving a subring, or what I have left to prove.

To prove this set is a ring, we must prove

(i) $S$ is closed under addition, and

(ii) $S$ is closed under multiplication, and

(iii)$S$ is associative on the operations, and

(iv) $S$ is commutative on the operations, and

(v) $S$ is distributive on the operations, and

(vi) $S$ has an additive identity ($0_R$), and

(vii) All operations in $S$ have an additive inverse.

We will begin by proving part (i). That is, we will prove $S$ is closed under addition. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and add it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})+(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then factor out $\sqrt[3]{2}$ and $\sqrt[3]{4}$ using the distributive property to get $(a+d+(b+e)\sqrt[3]{2}+(c+f)\sqrt[3]{4})$. Since $b,e,c,f \in \mathbb{Q}$, $(b+e),(c+f) \in \mathbb{Q}$, we know $S$ is closed under addition which proves part (i) of the proof.

We will now prove part (ii). That is, we will prove $S$ is closed under multiplication. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and multiply it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then multiply to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{4}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{4}+cd\sqrt[3]{4}+ce\sqrt[3]{2}\sqrt[3]{4}+cf\sqrt[3]{4}\sqrt[3]{4}$. We know $\sqrt[3]{4}=\sqrt[3]{2}\sqrt[3]{2}$, so we can make a substitution to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{2}\sqrt[3]{2}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cd\sqrt[3]{2}\sqrt[3]{2}+ce\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}$. We can then simplify this to $ad+ae\sqrt[3]{2}+af(2\sqrt[3]{2})+bd\sqrt[3]{2}+be(2\sqrt[3]{2})+bf(3\sqrt[3]{2})+cd(2\sqrt[3]{2})+ce(3\sqrt[3]{2})+cf(4\sqrt[3]{2})$ factor out $\sqrt[3]{2}$ using the distributive property to get $ad+(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4))\sqrt[3]{2}$. Since $a,b,c,d,e,f,2,3,4 \in \mathbb{Q}$, $(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4)) \in \mathbb{Q}$. Thus, we know $S$ is closed under multiplication which proves part (ii) of the proof.

We will now prove part (vi). That is, we will prove $S$ has an additive identity, or in other words has $0_R$. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$. Since $a,b,c \in \mathbb{Q}$ and $0 \in Q$, we will let $a$, $b$, and $c$ be $0$ to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}$. We can simplify this to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}=0+0+0=0_R$. Since the ring contains 0, we know it has an additive identity and proves part (vi) of the proof.

Can anyone help me get on track, please?

Answer 1

This is a subset of the real numbers with the same ring structure, you just have to see that it is closed under adition and multiplication as distribution, identity and associativity are free. Clearly, it is closed under addition. To prove that it is closed under adition we only have to check that products of (additive) generators are in the ring, Let $x = 2^{1/3}$, $x^2 = 4^{1/3}$ and $x^3 = 2 \in \mathbb{Q}$ so it is also closed under multiplication. Showing that it is a subring of the real numbers takes care of most of the proof.

Answer 2

Alternative proof: $\mathbb{Q}[x]$ is a ring and let $I$ be the ideal generated by $x^3 -2$, then we have that $R = \mathbb{Q}[x]/ I$ is a ring.

We have a homomorphism of rings $\phi: \mathbb{Q}[x] \to \mathbb{R}$ given by $\phi(1) = 1$ and $\phi(x) = \sqrt[3]{2}$, since $\phi(I) \subseteq \{0\}$ this defines a homomorphism $\phi: R \to \mathbb{R}$, now $R$ is generated by $1,x$ and the image of $\phi$ contains, $1, \sqrt[3]{2},\sqrt[3]{2} = \phi(x^2)$ (the image of a map is generated by generators of its domain) so in fact $S = \text{im}(\phi)$, as it is the image of a ring under a homomorphism of rings, it is a ring, a posteriori it is an isomorphism.