# Using First Isomorphism Theorm for Ring

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?

## 1 Answer

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 $f \in \langle x^3-2\rangle$.

• 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 – Hayden Apr 9 '17 at 14:58
• (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.) – Hayden Apr 9 '17 at 15:02
• Ah, noted. Thank you I just think I need to have that explained a bit more for me to grasp. – Sam Apr 9 '17 at 15:30
• Sure, if you have any other questions let me know – Hayden Apr 9 '17 at 15:31
• 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 $f(a+bx+cx^2)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. – Hayden Apr 9 '17 at 15:39