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Decide whether the following subset of $\mathbb{R}$ is not a ring, or is a ring but not a field, or is a field:

S = {${a + b\sqrt[3]{2} + c\sqrt[3]{4} | a,b,c \in \mathbb{Q} } $ }

I tried approaching by considering the polynomial $f(x) = x^3 - 2$ $\in \mathbb{Q}[x] $ and observing that a root is $x = \sqrt[3]{2}$. Then, if there exists a set containing $\mathbb{Q}$ and $\sqrt[3]{2}$, we prove that S is a field as S = $ \mathbb{Q}[\sqrt[3]{2}]$. However, I have been unsuccessful. Are there any hints for tackling this?

Anon
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  • is a ring, field, or neither --- Since fields are also rings, a clearer wording is: "is not a ring, or is a ring but not a field, or is a field". – Dave L. Renfro Feb 28 '24 at 03:26
  • Yes, I corrected it. Do you have any tips on approaching this? – Anon Feb 28 '24 at 03:47
  • The set is a field. Everything except the existence of inverses is straightforward, and to show inverses exist, one method that doesn't involve anything beyond school level algebra is to use a less trivial "rationalize the denominator" technique -- see [Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing](https://math.stackexchange.com/q/294993/13130) (especially my answer, which relies on a famous factorization formula). – Dave L. Renfro Feb 28 '24 at 04:57

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First show that $S$ is a ring by showing that it satisfies all ring properties.
Consider the map $\psi:\mathbb Q[x]\to S$ by sending $f(x)\to f(\sqrt[3]{2})$. It is clear that $\psi$ is surjective with kernel $(x^3-2)$. So by the first isomorphism theorem it follows that $S=\mathbb Q(\sqrt[3]{2})$.

ShyamalSayak
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  • We haven't seen that theorem yet in my course. Only theorems we can use are from Chapter 3.2 of Shifrin's "Abstract Algebra - A Geometric Approach". – Anon Feb 28 '24 at 03:49