# Minimal Polynomial

• Mar 4th 2010, 07:12 PM
chiph588@
Minimal Polynomial
Without finding the minimal polynomial for $\displaystyle r=i\cdot 2^{\frac{1}{3}}+i$, show the degree of said polynomial must be $\displaystyle 6$. I know this is true because I was able to derive $\displaystyle r\text{'s}$ minimal polynomial ($\displaystyle f(x)=x^6+3x^4-9x^2+9$), but in doing so I used the assumption that it was of degree $\displaystyle 6$ as opposed to $\displaystyle 2$ or $\displaystyle 3$.
• Mar 4th 2010, 08:03 PM
NonCommAlg
Quote:

Originally Posted by chiph588@
Without finding the minimal polynomial for $\displaystyle r=i\cdot 2^{\frac{1}{3}}+i$, show the degree of said polynomial must be $\displaystyle 6$. I know this is true because I was able to derive $\displaystyle r\text{'s}$ minimal polynomial ($\displaystyle f(x)=x^6+3x^4-9x^2+9$), but in doing so I used the assumption that it was of degree $\displaystyle 6$ as opposed to $\displaystyle 2$ or $\displaystyle 3$.

show that both $\displaystyle i$ and $\displaystyle \sqrt[3]{2}$ are in $\displaystyle \mathbb{Q}(i\sqrt[3]{2} + i)$ and thus $\displaystyle \mathbb{Q}(i\sqrt[3]{2} + i)=\mathbb{Q}(\sqrt[3]{2},i)$. we also have $\displaystyle [\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})] \times [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=2 \times 3 = 6.$
• Mar 4th 2010, 09:09 PM
chiph588@
Quote:

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

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

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

But is there and easier way to show $\displaystyle \sqrt[3]{2},i\in \mathbb{Q}(\alpha)$?
• Mar 4th 2010, 09:52 PM
NonCommAlg
Quote:

Originally Posted by chiph588@
Let $\displaystyle \alpha=i\sqrt[3]{2}+i$.

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

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

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