If I know a number, say $3 + \sqrt{-5}$ is an algebraic integer, it's easy to figure out the minimal polynomial: $x^2$ minus trace times $x$ plus norm, e.g., $x^2 - 6x + 14$.

But if the number is an algebraic number but not an algebraic integer, e.g., $$\frac{3 + \sqrt{-5}}{7},$$ how do I figure out the minimal polynomial? Without asking Wolfram Alpha?

Edit: Mark Bennett pointed out in a comment that the minimal polynomial formula I gave above is specifically for quadratic integers, not algebraic integers in general. I will consider as valid answers that address only algebraic numbers of degree $2$.

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    $\begingroup$ You do exactly the same thing.... $\endgroup$ – Lord Shark the Unknown Aug 31 '18 at 21:42
  • $\begingroup$ Alternatively, it's $\,(7x)^2 - 6 \cdot (7x) + 14\,$ down to a constant factor. $\endgroup$ – dxiv Aug 31 '18 at 21:43
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    $\begingroup$ Your method only works as described for algebraic integers with quadratic minimal polynomial - for higher degrees you need all the symmetric functions in the conjugates, not just the norm and trace. $\endgroup$ – Mark Bennet Aug 31 '18 at 21:48
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    $\begingroup$ But there’s nothing in your method that requires the irrationality in question to be integral. $\endgroup$ – Lubin Sep 4 '18 at 20:55
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    $\begingroup$ Wait, isn't every algebraic number times some integer an algebraic integer? $\endgroup$ – mathworker21 Sep 11 '18 at 17:27

(Since you're discussing minimal polynomials and algebraic numbers, I'm going to assume that you have some familiarity with Galois theory. If not, this solution might not be too helpful). The computation that I think you're interested in is as follows: In general, if you have some algebraic number, $\alpha$, you can find some Galois extension $K \supset \mathbb{Q}$ with $\alpha \in K$. Then the minimal polynomial of $\alpha$ is $m_\alpha(x) = \prod_{i} (x - \alpha_i)$ where the $\alpha_i$ are the distinct Galois conjugates of $\alpha$. The reasons this computation works are as follows:

You can see that the polynomial is guaranteed to have coefficients in $\mathbb{Q}$ because applying any element of $\operatorname{Gal}(K / \mathbb{Q})$ to $m_\alpha(x)$ will give the same polynomial. Hence, the coeffients of $m_\alpha(x)$ are fixed by all of the automorphisms in $\operatorname{Gal}(K / \mathbb{Q})$, implying that they are all in $\mathbb{Q}$.

Further, you can see the minimality of the degree of $m_\alpha$ with the following argument. Suppose that $f(x) \in \mathbb{Q}[x]$ has $f(\alpha) = 0$. Then for any $\sigma \in \operatorname{Gal}(K/\mathbb{Q})$, we have $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$, implying that every distinct Galois conjugate of $\alpha$ is a root of $f$. Hence, $m_\alpha(x)$ divides $f(x)$, which gives the minimality of its degree.

For the example you give above with $\alpha = \frac{3 + \sqrt{-5}}{7}$, we see that $\alpha \in \mathbb{Q}(\sqrt{-5})$, which has two members of its Galois group: the identity and the automorphism which fixes $\mathbb{Q}$ and sends $\sqrt{-5}$ to $-\sqrt{-5}$. Hence, the Galois conjugates of $\alpha$ are $\alpha$ itself and $\overline{\alpha} = \frac{3-\sqrt{-5}}{7}$. Then the minimal polynomial of $\alpha$ is given by $$m_\alpha(x) = (x - \alpha)(x - \overline{\alpha}) = x^2 - \frac{6}{7}x + \frac{2}{7}$$


The method I use for doing it is quite different from that of Greg.

As Greg does, I find a field $K$ containing the quantity $\alpha$, and one needs to know a $\Bbb Q$-basis of $K$, say $\{u_1,u_2,\cdots,u_n\}$. Then I find the characteristic polynomial of the $\Bbb Q$-linear map $\rho_\alpha:K\to K$, defined by $z\mapsto\alpha z$. (This is the “regular representation” in action.) You use the basis for this, of course, and the task can be tedious. In case you have been so lucky as to take $K=\Bbb Q(\alpha)$, then this characteristic polynomial will be what you’re looking for. Otherwise, you’ll get a power of the minimal polynomial; and which case holds can easily be checked.

  • $\begingroup$ Maybe I should have said too that if you know the degree of $\alpha$, say $n$, then the basis $\{1,\alpha,\cdots,\alpha^{n-1}\}$ gives a particularly nice matrix. $\endgroup$ – Lubin Sep 5 '18 at 0:55

I just want to elaborate on a comment, so I'm marking this community wiki.

You do exactly the same thing...

At least this is true for quadratic numbers. So, like you said,

$x^2$ minus trace times $x$ plus norm,

so that in the case of $$\frac{3 + \sqrt{-5}}{7},$$ we have $$x^2 - \frac{6x}{7} + \frac{14}{49} = x^2 - \frac{6x}{7} + \frac{2}{7}.$$

Only problem is that not all the coefficients are integers, though they are rational. How about multiplying the whole thing by $7$ to obtain $7x^2 - 6x + 2$?

Just to verify this is correct, go to Wolfram|Alpha and put in roots of 7x^2 - 6x + 2. It should answer $$x = \frac{1}{7}(3 - i \sqrt{5})$$ $$x = \frac{1}{7}(3 + i \sqrt{5})$$


The trace/norm formula for the minimal polynomial of a quadratic irrational is valid for all algebraic numbers, not just algebraic integers, as seen in the example in the discussion—this is because we can express the trace and norm in terms of sums and products of images of our element over all embeddings of the quadratic number field in the complex numbers.

The discussion contains essentially everything there is to say about minimal polynomials, which are a typical example of an object with a conceptual definition (smallest degree monic polynomial with a given zero), a reformulation useful for proofs (the expression in terms of the Galois group), and a reformulation useful for computation (characteristic polynomial of multiplication by $\alpha$ on $\mathbb{Q}(\alpha)$).

In general there is not any procedure more efficient than the one described in the discussion, using linear algebra. In some simple examples there are some tricks that can be used. For example, suppose we have a tower of extensions of number fields $L / K / \mathbb{Q}$, and $L = \mathbb{Q}(\alpha)$ for some algebraic number $\alpha$. It might be easier to first calculate the minimal polynomial of $\alpha$ over $K$, and then take the product of the conjugates of this polynomial to get something with $\mathbb{Q}$ coefficients.

We can try this in the simple example $L = \mathbb{Q}(i, \sqrt{2})$, $K = \sqrt{2}$, $\alpha = i + \sqrt{2}$. Note, however, that this boils down in the end to expressing the minimal polynomial as a product of the $(X - \alpha')$, where $\alpha'$ runs over the conjugates of $\alpha$.


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