# General way to determine $\mathbb{Q}(\gamma) = \mathbb{Q}(\alpha,\beta)$ given $\alpha$ and $\beta$

I'm currently reading S. Lang's "Undergraduate Algebra". After the primitive root element theorem (Field theory chapter), there are a bunch of exercises to find one primitive element of extensions and, then, their degrees. However, I don't even know how I should start. They are as below:

1. In each case find an element $$\gamma$$ such that $$\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\gamma)$$. Prove every statement you make.

a) $$\alpha = \sqrt{-5}, \beta = \sqrt{2}$$

b) $$\alpha = \sqrt{2} , \beta = \sqrt{2}$$

c) $$\alpha =$$ root of $$t^3 -t + 1$$ , $$\beta =$$ root of $$t^2-t-1$$

d) $$\alpha =$$ root of $$t^3 -2t + 3$$, $$\beta =$$ root of $$t^2 + t + 2$$

$$\quad$$2. Find the degrees of the fields $$\mathbb{Q}(\alpha, \beta)$$ over $$\mathbb{Q}$$ in each case of exercise 1.

I think that exercises a) and b) proceed pretty much the same way, but I'm not sure about c) and d).

• I solved a), but I'm not sure if my approach is correct. I first showed that the degree of $\mathbb{Q}(\sqrt{2},\sqrt{-5})$ = degree of $\mathbb{Q}(\sqrt{2}+\sqrt{-5})$ = 4. Since this second field is a subspace of the first, with same dimension, they must be equal. Note that I had to solve 2a) before 1a), so this probably isn't the easier way to solve it. Dec 20, 2019 at 2:39
• Have you checked the wiki page en.wikipedia.org/wiki/Primitive_element_theorem Dec 20, 2019 at 3:07
• Yes. I didn’t understand the construction, and how I can’t find that $\sqrt{2} + \sqrt{-5}$ is a primitive element without calculating all its powers. Dec 20, 2019 at 3:12
• In Milne's notes on field theory, he explicitly determines the set of $c \in F$ such that $\gamma := \alpha + c \beta$ is a primitive element for $F(\alpha, \beta)$. See the proof of Theorem 5.1, specifically the paragraph at the top of p. 60. Dec 20, 2019 at 15:43

So let me use exercise c) to illustrate the idea.

In general, if $$\alpha$$ and $$\beta$$ are two algebraic numbers, then for all but finitely many rational numbers $$x$$, the element $$\alpha + x\beta$$ is a primitive element of $$\Bbb Q(\alpha, \beta)$$.

So let's simply look for such elements $$\gamma$$ of the form $$\alpha + x\beta$$ with $$x\in\Bbb Q$$.

In exercise c), we have:

• $$\alpha ^ 3 -\alpha + 1 = 0$$;
• $$\beta^2 - \beta - 1 = 0$$;
• $$\{1, \alpha, \alpha^2\}$$ is a basis of $$\Bbb Q(\alpha)/\Bbb Q$$;
• $$\{1, \beta\}$$ is a basis of $$\Bbb Q(\beta)/\Bbb Q$$.

Hence we get a basis of $$\Bbb Q(\alpha, \beta)/\Bbb Q$$, which is simply $$\{1, \alpha, \alpha^2, \beta, \alpha\beta, \alpha^2\beta\}$$.

By definition, an element $$\gamma = \alpha + x\beta$$ is a primitive element of $$\Bbb Q(\alpha, \beta)/\Bbb Q$$ if and only if $$\{1, \gamma, \gamma^2, \gamma^3, \gamma^4, \gamma^5\}$$ is a basis of $$\Bbb Q(\alpha, \beta)/\Bbb Q$$. Since we already have a basis, we may write every element as $$\Bbb Q$$-linear combination of this basis: $$\begin{eqnarray*} 1 &=& 1\times 1 + 0 \times \alpha + 0 \times \alpha^2 + 0 \times \beta + 0 \times \alpha\beta + 0 \times \alpha^2\beta\\ \gamma &=& 0 \times 1 + 1 \times \alpha + 0 \times \alpha^2 + x \times \beta + 0 \times \alpha\beta + 0\times\alpha^2\beta\\ \gamma^2 &=& x^2 \times 1 + 0 \times \alpha + 1 \times \alpha^2 + x^2 \times \beta + 2x \times \alpha\beta + 0\times\alpha^2\beta\\ \gamma^3 &=& (x^3 - 1) \times 1 + (3x^2 + 1) \times \alpha + 0 \times \alpha^2 + 2x^3 \times \beta + 3x^2 \times \alpha\beta + 3x\times\alpha^2\beta\\ \gamma^4 &=& 2x^4 \times 1 + (4x^3 - 1) \times \alpha + (6x^2 + 1) \times \alpha^2 + (3x^4 - 4x) \times \beta + (8x^3 + 4x) \times \alpha\beta + 6x^2\times\alpha^2\beta\\ \gamma^5 &=& (3x^5 - 10x^2 - 1) \times 1 + (10x^4 + 10x^2 + 1) \times \alpha + (10x^3 - 1) \times \alpha^2 + (5x^5 - 10x^2) \times \beta + (15x^4 + 10x^2 - 5x) \times \alpha\beta + (20x^3 + 5x)\times\alpha^2\beta \end{eqnarray*}$$

To get the above identities, we just keep multiplying the previous line by $$\gamma$$ and using the relations $$\alpha^3 = \alpha - 1$$ and $$\beta^2 = \beta + 1$$.

Written in matrix form, this becomes:

$$(1, \gamma, \gamma^2, \gamma^3, \gamma^4, \gamma^5) = (1, \alpha, \alpha^2, \beta, \alpha\beta, \alpha^2\beta)\cdot M,$$

where $$M$$ is the following matrix: $$\begin{pmatrix} 1 & 0 & x^2 & x^3 - 1 & 2x^4 & 3x^5 - 10x^2 - 1\\ 0 & 1 & 0 & 3x^2 + 1 & 4x^3 - 1 & 10x^4 + 10x^2 + 1\\ 0 & 0 & 1 & 0 & 6x^2 + 1 & 10x^3 - 1\\ 0 & x & x^2 & 2x^3 & 3x^4 - 4x & 5x^5 - 10x^2\\ 0 & 0 & 2x & 3x^2 & 8x^3 + 4x & 15x^4 + 10x^2 - 5x\\ 0 & 0 & 0 & 3x & 6x^2 & 20x^3 + 5x \end{pmatrix}$$ Therefore, $$\gamma$$ is a primitive element if and only if the matrix $$M$$ is invertible, i.e. the determinant is non-zero.

Calculation shows that $$\det(M) = 125x^9 - 150x^7 + 45x^5 + 23x^3$$. Thus we may take e.g. $$x = 1$$ and get that $$\alpha + \beta$$ is a primitive element (in fact, the only rational root of this polynomial being $$x = 0$$, we see that $$\alpha + x\beta$$ is a primitive element for any $$x \neq 0$$).

Some clarifications:

• Why did I bother keeping the $$x$$ as a variable during all the calculations? Wouldn't it be simpler to replace $$x$$ by $$1$$ everywhere?

Yes, it would be much simpler and the calculation would look less tedious. But what if $$\alpha + \beta$$ happens to be non-primitive? Since we don't know a priori which $$x$$ gives a primitive element, I tend to keep it as a variable so that we can easily choose it in the very last step.

• How did I do such a complicated computation?

With a computer. This method, although complicated, is easily automated. It's quite simple to implement the algorithm with the help of some computer algebra system.

• Are there simpler methods?

Sometimes yes. But most simpler methods are usually only applicable to certain cases, so they are less "universal". Also, you might need knowledge of deeper mathematics.

This method, however, is of algorithmic nature, applicable (at least) to all field extensions of characteristic $$0$$, and only requires basic linear algebra.

• I understood the part above the first horizontal line, understood that in $\alpha$ is algebraic of degree n then $\mathbb{Q}(\alpha)/\mathbb{Q}$ is generated by $\{1, \alpha, \ldots, \alpha^{n-1}\}$. But why is $\{1, \alpha, \alpha^2, \beta, \alpha \beta, \alpha^2\beta\}$ a basis to $\mathbb{Q}(\alpha, \beta)/\mathbb{Q}$? Dec 20, 2019 at 4:14
• You first need to see that the extension $\Bbb Q(\alpha, \beta)/\Bbb Q$ has degree $6$. Once that is done, it suffices to see that every element in $\Bbb Q(\alpha, \beta)$ can be written as $\Bbb Q$-linear combination of these $6$ elements. Do you have difficulty in either part? Dec 20, 2019 at 4:18
• Both. I see that $n \,:= [\mathbb{Q}(\alpha, \beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$ so $2,3| n \implies 6|n,$ but I can't directly show this. Dec 20, 2019 at 4:24
• $\Bbb Q(\alpha, \beta)$ is obtained from $\Bbb Q(\alpha)$ by adding a root of $t^2 - t - 1$, so it's either equal to $\Bbb Q(\alpha)$ (when $\Bbb Q(\alpha)$ already contains the root) or is an extension of degree $2$ of $\Bbb Q(\alpha)$ (when $t^2 - t - 1$ is irreducible over $\Bbb Q(\alpha)$). The former case is impossible, as you noticed, because the degree of $\Bbb Q(\alpha, \beta)$ cannot be $3$: it must be divisible by $2$. Hence we have $[\Bbb Q(\alpha, \beta) :\Bbb Q(\alpha)] = 2$ and therefore its degree over $\Bbb Q$ is $6$. Dec 20, 2019 at 4:28
• In general, if $(e_i)_i$ is a basis of $K/F$ and $(f_j)_j$ is a basis of $L/K$, then $(e_if_j)_{i, j}$ is a basis of $L/F$. Here we have $L = \Bbb Q(\alpha, \beta)$, $K = \Bbb Q(\alpha)$, $F = \Bbb Q$. Dec 20, 2019 at 4:38