Minimal Polynomial

**chiph588@** · March 4th 2010, 07:12 PM

Without finding the minimal polynomial for $r=i\cdot 2^{\frac{1}{3}}+i$ , show the degree of said polynomial must be $6$ . I know this is true because I was able to derive $r\text{'s}$ minimal polynomial ( $f(x)=x^6+3x^4-9x^2+9$ ), but in doing so I used the assumption that it was of degree $6$ as opposed to $2$ or $3$ .

**NonCommAlg** · March 4th 2010, 08:03 PM

Originally Posted by chiph588@

Without finding the minimal polynomial for $r=i\cdot 2^{\frac{1}{3}}+i$ , show the degree of said polynomial must be $6$ . I know this is true because I was able to derive $r\text{'s}$ minimal polynomial ( $f(x)=x^6+3x^4-9x^2+9$ ), but in doing so I used the assumption that it was of degree $6$ as opposed to $2$ or $3$ .

show that both $i$ and $\sqrt[3]{2}$ are in $\mathbb{Q}(i\sqrt[3]{2} + i)$ and thus $\mathbb{Q}(i\sqrt[3]{2} + i)=\mathbb{Q}(\sqrt[3]{2},i)$ . we also have $[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})] \times [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=2 \times 3 = 6.$

**chiph588@** · March 4th 2010, 09:09 PM

Originally Posted by NonCommAlg

show that both $i$ and $\sqrt[3]{2}$ are in $\mathbb{Q}(i\sqrt[3]{2} + i)$ and thus $\mathbb{Q}(i\sqrt[3]{2} + i)=\mathbb{Q}(\sqrt[3]{2},i)$ . we also have $[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})] \times [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=2 \times 3 = 6.$

Let $\alpha=i\sqrt[3]{2}+i$ .

It turns out $\sqrt[3]{2} = -\frac{1}{6}\alpha^4-\alpha^2+\frac{1}{2}$
and $i = \frac{1}{6}\alpha^5+\frac{2}{3}\alpha^3-\frac{1}{2}\alpha$ ,
so indeed we know $\mathbb{Q}(\sqrt[3]{2},i) \subseteq \mathbb{Q}(\alpha)$ .

But is there and easier way to show $\sqrt[3]{2},i\in \mathbb{Q}(\alpha)$ ?

**NonCommAlg** · March 4th 2010, 09:52 PM

Originally Posted by chiph588@

Let $\alpha=i\sqrt[3]{2}+i$ .

It turns out $\sqrt[3]{2} = -\frac{1}{6}\alpha^4-\alpha^2+\frac{1}{2}$
and $i = \frac{1}{6}\alpha^5+\frac{2}{3}\alpha^3-\frac{1}{2}\alpha$ ,
so indeed we know $\mathbb{Q}(\sqrt[3]{2},i) \subseteq \mathbb{Q}(\alpha)$ .

But is there and easier way to show $\sqrt[3]{2},i\in \mathbb{Q}(\alpha)$ ?

well, this is how i did it: $\alpha^3=-3i(1+\sqrt[3]{2} + \sqrt[3]{4})=-3 \alpha - 3i \sqrt[3]{4}$ and so $\beta=i \sqrt[3]{4} \in \mathbb{Q}(\alpha)$ and we're done because $\sqrt[3]{2}=\frac{-\beta^2}{2}$ and $i=\frac{-\beta^3}{4}.$

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