I am asked to prove that $\mathbb{Q}[x]/(x^3-2)\cong R$, where $R$ is the set of numbers $a+b\sqrt[3]2+c\sqrt[3]4$ with a,b,c $\in\mathbb{Q}$.

By, first isomorphism theorem of rings:

I have defined my map as $\mathbb{Q}[x]\rightarrow R$ by $(f(x))$$\rightarrow f(\sqrt[3]2)$.

I am not sure how to show surjective or how to go about the kernel. Would proving containment be best for the kernel?


Your map $\theta$ is a good choice. For surjectivity, note that $f(x) = a + bx + cx^2$ is mapped to $a+b\sqrt[3]{2}+c\sqrt[3]{4}$.

Your intuition for how to prove that $\ker \theta = \langle x^3-2\rangle$ is also good.

By definition of $\theta$, $\theta( x^3-2) = \sqrt[3]{2}^3 - 2 = 0$, so $\langle x^3-2\rangle \subset \ker \theta$.

For the other direction, if $f(x) \in \ker \theta$, then $f(\sqrt[3]{2})=0$. As such, $\sqrt[3]{2}$ is a root of $f$, and so the minimal polynomial of $\sqrt[3]{2}$ must divide $f$. But that minimal polynomial is $x^3-2$, and $(x^3-2) | f$ is exactly what it means to say that $f \in \langle x^3-2\rangle$.

  • $\begingroup$ Yes, that's exactly what surjectivity is for $\theta$; for any element $r$ in $R$, there is some polynomial $f(x) \in \mathbb{Q}[x]$ such that $\theta(f(x)) = r$. $x^3-2$ is the minimal polynomial of $\sqrt[3]{2}$ because it is a polynomial with root $\sqrt[3]{2}$, and no polynomial of strictly smaller degree (in $\mathbb{Q}[x]$) has $\sqrt[3]{2}$ as a root $\endgroup$ – Hayden Apr 9 '17 at 14:58
  • $\begingroup$ (if $x^3-2$ wasn't the minimal polynomial, we would still have that the minimal polynomial of $\sqrt[3]{2}$ would need to divide $x^3-2$, but then $x^3-2$ would need to split into a linear and a quadratic factor in $\mathbb{Q}$, which is not possible because the roots of $x^3-2$ are not rational.) $\endgroup$ – Hayden Apr 9 '17 at 15:02
  • $\begingroup$ Ah, noted. Thank you I just think I need to have that explained a bit more for me to grasp. $\endgroup$ – Sam Apr 9 '17 at 15:30
  • $\begingroup$ Sure, if you have any other questions let me know $\endgroup$ – Hayden Apr 9 '17 at 15:31
  • $\begingroup$ In general, a function $f: A \to B$ is surjective if for every $b\in B$ there exists $a\in A$ such that $f(a)=b$. So you'd start by taking an element $a+b\sqrt[3]{2} + c\sqrt[3]{4} \in R$, then observe that $f(a+bx+cx^2)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. $\endgroup$ – Hayden Apr 9 '17 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.