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I need to find a subring of $\mathbb{ R }$ isomorphic to $\mathbb{ Q }[ x ] / \langle x^3 - 2 \rangle$.

I considered using the Isomorphism Theorem to try to find a homomorphism $\theta: \mathbb{ Q }[x] \to R \subseteq \mathbb{ R }$ such that $\langle x^3 - 2 \rangle = \ker \theta$, so then $R \cong \mathbb{ Q }[ x ] / \langle x^3 - 2 \rangle$. But I was unsure of where to proceed.

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The domain of $\theta$ is the polynomial ring $\Bbb Q[x]$ instead of $\Bbb Q$. – Berci Nov 30 '12 at 16:23
up vote 9 down vote accepted

You want $x^3-2$ to be in $\ker \theta$, i.e. $\theta(x^3-2)=0$. Since $\theta$ is a homomorphism, this implies that $\theta(x)^3-2=0$. Can you think of a real number $y$ such that $y^3-2=0$?

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So then $y = \sqrt[3]{2}$. Oh I see now, thanks! – Carlos Nov 30 '12 at 16:27

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