# Minimal polynomials of $\sqrt{2}i$ over $\mathbb{Q}$ and $\mathbb{R}$

Can someone tell me if this is right:

I would like to find the minimal polynomial of

(i) $\sqrt{2}i$ over $\mathbb{Q}$

(ii) $\sqrt{2}i$ over $\mathbb{R}$

(i):

$\sqrt{2}i$ is a root of $f(x) = x^4-2$. This is already monic, so to show that this is a minimal polynomial I only need to show that it is irreducible. Edit: To do that, I can use the Eisenstein: $p=2$ does not divide $a_1 = 1$ and $p^2$ does not divide $a_0 = -2$, therefore it is irreducible over $\mathbb{Q}$.

(ii):

This time, $\sqrt{2}i$ is a root of $f(x) = x^2+\sqrt{2}$. For polynomials of degree two it's enough to check if they have a root. The only roots this one has are complex therefore it is irreducible over $\mathbb{R}$ and therefore the minimal polynomial.

So my more general question is: is the way to find a minimal polynomial of an element $a$ in general to find a polynomial $f$ such that $f(a) = 0$ and then to norm $f$ and then to show that $f$ is irreducible?

• Part (i) is incorrect: lack of roots does not imply irreducibility when the polynomial is of degree greater than 3, because the polynomial could split into a product of two irreducible polynomials, each of degree $2$. You have shown $x^4-2$ does not have a linear factor modulo $3$, but could be the product of two quadratics? I would instead suggest using Eisenstein's Criterion on $x^4-2$. (ii) is okay; for your final question, a top-down approach is what you suggest, except you can just try to find an irreducible factor of $f$ that has $a$ as a root. Feb 13, 2011 at 19:55
• I'm pretty sure that $f(x) = (x^2+x+2)(x^2+2x+2)$ modulo $3$ Feb 13, 2011 at 20:01
• Many thanks to both of you. Arturo, maybe you could copy paste your comment into an answer so that I can accept it as the answer to this question? Feb 13, 2011 at 20:16

In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.

For example, let $\alpha=\sqrt{2} + \sqrt{3}$. Then \begin{align*} \alpha^2 &= 5+2\sqrt{6} \\\ \alpha^3 &= 11\sqrt{2}+9\sqrt{3} \\\ \alpha^4 &= 49+20\sqrt{6} \end{align*} Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors: $$1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix}, \qquad \alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix}, \qquad \alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 2\end{bmatrix}, \qquad \alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix}, \qquad \alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix}$$ As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial. However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with $$\alpha^4 - 10\alpha^2 + 1 \;=\; 0$$ It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.

This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$. Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$. In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.

For example, if $\alpha = i\sqrt{2}$, then $$\alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt{8},\qquad \alpha^4 = 2.$$ It seems clear that $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$. Thus $x^4-2$ should indeed be the minimum polynomial over $\mathbb{Q}$, though the easiest way to prove it is to show that $x^4 -2$ is irreducible. This can be done using Eisenstein's criterion, or by checking that $x^4-2$ has no roots and does not factor into quadratics modulo $4$.

• Thank you, your answer is very helpful. I just started to learn Galois theory, although I don't yet see how to use it to prove linear independence of elements in a field extension. What I know so far is what a Galois group is. Feb 13, 2011 at 21:14
• Jim, Why are $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ linearly independent? over $\mathbb{Q}$ Mar 28, 2011 at 12:50
• Yes, they are. Suppose $\sqrt 2$ is in $Q(\sqrt3)$, then $\sqrt 2= a+bsqrt3$. $2= a^2+3+2ab \sqrt 3$. Hence $\sqrt 3 = \frac{-a^2-1}{2ab}$ but the left hand side is rational number while $\sqrt 3$ is not. Contradiction. Apply the same to show they are all linearly independent Apr 29, 2015 at 20:09

Part (i) as given is incorrect: lack of roots does not imply irreducibility when the polynomial is of degree greater than 3, because the polynomial could split into irreducible polynomials of degrees greater than 1.

You have shown that $x^4-2$ has no linear factors modulo $3$, but you cannot conclude from this that it is irreducible modulo $3$: it could have two irreducible quadratic factors. And indeed, as chandok writes, \begin{align*} (x^2+x+2)(x^2+2x+2) &= x^4 + 2x^3 + 2x^2 + x^3 + 2x^2 + 2x + 2x^2 + 4x + 4\\ &= x^4 + 3x^3 + 6x^2 + 6x + 4 = x^4 + 1 = x^4 - 2, \end{align*} so the polynomial is not irreducible over $\mathbb{F}_3$.

Instead, I would suggest using Eisenstein's Criterion to show $x^4-2$ is irreducible over $\mathbb{Q}$, as it is pretty easy to apply there.

Part (ii) is correct.

As to your final question: there is a top-down approach and a bottoms-up approach. In the top down approach, if you can find any polynomial $f(x)$ that has $a$ as a root, then you know that the minimal polynomial will divide $f(x)$; in fact, it will be an irreducible factor of $f$. So you can try to find which irreducible factor of $f$ has $a$ as a root. This may be $f$ itself, or some factor. For example, to find the minimal polynomial of a $7$th root of unity, you could use the fact that it satisfies $x^7-1$. Then factor $x^7-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$, and you know that $a$ is a root of either $x-1$ or the second factor. Then you would look at the second factor (since presumably your $a$ is not $1$). And so on.

The bottoms-up approach is described by Jim Belk in his answer, where you "build up" the polynomial by considering powers of $a$ (or alternatively, images under the action of some Galois group).