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Let $\mathbb{Q}$ be the field of rational number. Suppose that $\alpha$ is a root of an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Q}[x]$. Then $\mathbb{Q}[\alpha]$ is the field extension of $\mathbb{Q}$ and $\mathbb{Q}[\alpha]\cong \mathbb{Q}[x]/(f(x))$.

My question is: How can I represent an arbitrary element of $\mathbb{Q}[\alpha]$ in term of $\alpha$ and some elements in $\mathbb{Q}$?

Is : $\mathbb{Q}[\alpha]=\lbrace a_{0}+a_1\alpha+...+a_{n-1}\alpha^{n-1}|\ \text{for some}\ a_{i}\in \mathbb{Q} \rbrace$?

I have a concrete problem with $\mathbb{Q}[\sqrt[3]{2}\zeta]$ where $\zeta \neq 1$ is the third root of unity, I do not know how to represent exactly an arbitrary element of $\mathbb{Q}[\sqrt[3]{2}\zeta]$ in term of elements in $\mathbb{Q}$ and $\sqrt[3]{2}, \zeta$.

Please help me.

Thanks.

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  • $\begingroup$ The answer to your question about $\Bbb Q[\alpha]$ is yes. Can you tell why? This subsequently answers your next question about $\Bbb Q[\sqrt[3]{2}\zeta]$. $\endgroup$ – anon Jan 3 '14 at 11:25
  • $\begingroup$ $x$ is a root of $f(x)$ in $\mathbb{Q}[x]/(f(x))$. This should make the rest clear. $\endgroup$ – Gina Jan 3 '14 at 11:26
  • $\begingroup$ Your first answer is exactly correct. So letting $\alpha = \sqrt[3]{2}\zeta$, since $\alpha$ is a root of $X^3 - 2$, every element of $\mathbf{Q}(\alpha)$ can be represented uniquely as $a + b\alpha + c\alpha^2$ for $a, b, c \in \mathbf{Q}$. Multiplication of elements is done in the expected way, taking into account the relation $\alpha^3 = 2$. $\endgroup$ – user118829 Jan 3 '14 at 11:28
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Yes your element description of $\mathbb{Q}[\alpha]$ is correct. For $\alpha = 2^{1/3} \cdot \zeta_3$ we have $\alpha^3=2$, and hence $f=x^3-2$ (this is irreducible by Eisenstein). Thus $\mathbb{Q}[\alpha]$ has as $\mathbb{Q}$-basis $1,\alpha,\alpha^2$.

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  • $\begingroup$ So elements of $\mathbb{Q}[/alpha]$ are represented by $a+b\alpha+c\alpha^{2}$ where $\alpha=\sqrt[3]{2}\zeta$, and $a,b,c \in \mathbb{Q}$ ? $\endgroup$ – Arsenaler Jan 3 '14 at 11:31
  • $\begingroup$ Yes. $\phantom{ }$ $\endgroup$ – Martin Brandenburg Jan 3 '14 at 11:34

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