Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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

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


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

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

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

share|cite|improve this answer

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.

share|cite|improve this answer
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, $\mathbb{Q}(\alpha,i)=\mathbb{Q}(\alpha+i)$.

share|cite|improve this answer

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!

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.