# Write down a $\mathbb Q$-Basis for $\mathbb Q(\alpha, i)$ and show that $\mathbb Q(\alpha, i) = \mathbb Q(\alpha + i)$.

I have found a $\mathbb Q$-basis for $\mathbb Q(\alpha)$, where $\alpha$ is a root of $x^{3}-x+1$, to be $\{1, \alpha, \alpha^{2}\}$ & a $\mathbb Q(\alpha)$-basis for $\mathbb Q(\alpha, i)$ to be $\{1, i\}$. Hence a $\mathbb Q$-basis for $\mathbb Q(\alpha, i)$ is $\{1, \alpha, \alpha^{2}, i, i\alpha, i\alpha^{2}\}$.

Now I have no idea how to show $\mathbb Q(\alpha, i) = \mathbb Q(\alpha + i)$ or indeed to do that for any algebraic field extensions?

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Hint: It is sufficient to show that both fields have the same basis as a $\mathbb Q$-vector space. –  Johannes Kloos May 11 '12 at 17:48

Since $\alpha + i \in \mathbb{Q}(\alpha,i)$, we have $\mathbb{Q}(\alpha+i) \subset \mathbb{Q}(\alpha,i)$. Now to show the other inclusion, you need to show $i \in \mathbb{Q}(\alpha+i)$ (from which you also get $\alpha \in \mathbb{Q}(\alpha+i)$). To make the computation clearer, denote $\alpha + i = \beta$. From $\alpha^3 - \alpha + 1 = 0$, we get

$$(\beta-i)^3 - (\beta-i) + 1 = \beta^3 - 3i \beta^2 - 2 \beta + 1 = 0$$

So

$$\beta^3 - 2 \beta + 1 = 3i \beta^2$$

The right-hand side is non zero, so the left-hand side too, and we get

$$i = \frac{\beta^3 - 2 \beta + 1}{3 \beta^2} \in \mathbb{Q}(\beta)$$

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It is clear that $\mathbb{Q}(\alpha+i) \subseteq \mathbb{Q}(\alpha, i)$. You could try to show that $\alpha\in\mathbb{Q}(\alpha+i)$. Then $i\in\mathbb{Q}(\alpha+i)$ and so $\mathbb{Q}(\alpha,i)\subseteq\mathbb{Q}(\alpha+i)$ thereby solving the question.

Normally, you would try to find some expression involving only addition, subtraction, multiplication, division, rational numbers and $\alpha+i$ that equals $\alpha$ (or $i$, as Joel Cohen did). There is also an algorithmic way to check whether $\alpha\in\mathbb{Q}(\alpha+i)$. It is basically a structured way of finding such an expression.

Write down $(\alpha+i)^k$ in coordinates with respect to the basis you found. For example, if $k=3$, you get $$(\alpha+i)^3 = \alpha^3+3i\alpha^2 - 3\alpha - i = 3i\alpha^2-i-1-2\alpha$$ where I have used the relation $\alpha^3-\alpha+1=0$. Let us (for clarity) write this as a 6-tuple of coordinates with respect to $\{1,\alpha,\alpha^2,i,i\alpha,i\alpha^2\}$. We get $$(\alpha+i)^3 \cong (-1,-2,0,-1,0,3).$$ Doing this not only for $k=3$, but for $k\in\{0,\ldots,5\}$, you get six tuples of coordinates. The question whether $\alpha\in\mathbb{Q}(\alpha+i)$ then comes down to: is there a linear combination of these tuples giving $(0,1,0,0,0,0)\cong\alpha$?

If you find such a linear combination (in this case you will, because the six vectors will be linearly independent hence span all of $\mathbb{Q}^6$), you have found a linear combination of powers of $(\alpha+i)$ (i.e. a polynomial expression in $(\alpha+i)$) that equals $\alpha$. This of course guarantees that $\alpha\in\mathbb{Q}(\alpha+i)$, and you're done. I admit this takes some calculations, but it is a surefire way of solving the question.

Remark. There is a nice link with a common proof of the primitive element theorem. It is proved by finding $\lambda$ such that $\mathbb{Q}(a,b)=\mathbb{Q}(a+\lambda b)$. In the proof, it turns out that this will be true for all but finitely many values for $\lambda$, so typically we have that $\mathbb{Q}(a,b)=\mathbb{Q}(a+\lambda b)$. This is the same as saying that $n$ different vectors in $\mathbb{Q}^n$ will typically span all of $\mathbb{Q}^n$. I have never checked whether this link can be made precise, so it's just the intuition I have about this.

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Very helpful intuitively, thank you. –  joeF May 11 '12 at 21:40

Here's a purely naive approach, using the independences that you already know.

Note that $1,(\alpha+i),(\alpha+i)^2,(\alpha+i)^3$ are $\mathbb{Q}$-linearly independent. Indeed, \begin{align*} (\alpha+i)^2 &= (\alpha^2 -1) + 2i\alpha\\ (\alpha+i)^3 &= (\alpha^3 - 3\alpha) + i(3\alpha^2-1)\\ &= (-2\alpha-1) + i(3\alpha^2-1) \end{align*} (using the fact that $\alpha^3 = \alpha-1$). If $$a + b(\alpha+i) + c(\alpha+i)^2 + d(\alpha+i)^3 = 0,\quad a,b,c,d\in\mathbb{Q}$$ we have, looking at the real part, that $$(a-c-d) + (b-2d)\alpha + c\alpha^2 = 0.$$ Since $1,\alpha,\alpha^2$ are $\mathbb{Q}$-linearly independent, it follows that $c=0$, $b=2d$, and $a-c-d=0$. Looking at the imaginary part, we have $$b-d + 2c\alpha + 3d\alpha^2 = 0,$$ so again, using the independence of $1,\alpha,\alpha^2$, we conclude that $d=0$; hence $b=0$, and so $a=0$.

Thus, $[\mathbb{Q}(\alpha+i):\mathbb{Q}]\geq 4$. Since $\mathbb{Q}(\alpha+i)\subseteq \mathbb{Q}(\alpha,i)$, then by Dedekind's product theorem we have $$6 = [\mathbb{Q}(\alpha,i):\mathbb{Q}] = [\mathbb{Q}(\alpha,i):\mathbb{Q}(\alpha+i)][\mathbb{Q}(\alpha+i):\mathbb{Q}].$$ So $[\mathbb{Q}(\alpha+i):\mathbb{Q}]$ must divide $6$ and is at least $4$. Therefore, $[\mathbb{Q}(\alpha+i):\mathbb{Q}]=6$, and $[\mathbb{Q}(\alpha,i):\mathbb{Q}(\alpha+i)] = 1$. Therefore, $\mathbb{Q}(\alpha,i)=\mathbb{Q}(\alpha+i)$.

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There are lots of ways to do this. The easiest way is to show that the degree of each extension over $\mathbb{Q}$ is the same, and since $\mathbb{Q}(\alpha+i)$ is obviously in $\mathbb{Q}(\alpha,i)$ they must be the same field. Another way is to use the primitive element theorem by showing that $\alpha+i$ is not in any of the intermediate field extensions and therefore the element generates the whole extension.

I highly recommend Cox's book for this stuff!

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Do you mean David A. Cox Galois Theory? Will be doing Galois Theory next semester, this could be useful. Thank you. –  joeF May 11 '12 at 21:38
@joeF yes, it's the best book on the subject...the first chapter is kinda him trying too hard to motivate it and it's mostly a computational mess, but after that it's amazing. –  Steven-Owen May 11 '12 at 22:52

A method that works for this specific extension and is not at all general: look locally, at the prime $3$, and see that the first extension has residue-field extension degree $3$, while the second has rfd equal to $2$. But even over ${\mathbb{F}}_3$, $\alpha+i$ generates the field ${\mathbb{F}}_{3^6}$. Since the rfd of ${\mathbb{Q}}(\alpha+i)$ over $\mathbb{Q}$ is $6$, the field extension degree is at least that large.

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