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I am trying to find Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$. I was able to determine the minimal polynomial is fourth degree with roots at $i-\sqrt{2}$, $i+\sqrt{2}$,$-i-\sqrt{2}$,$-i+\sqrt{2}$. However I got this answer by guessing at what the roots should be. Is there a general technique for this type of problem.

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up vote 4 down vote accepted

First, note that $i+\sqrt{2}\in\mathbb{Q}(i,\sqrt{2})$, so the degree is either $2$ or $4$. But $i+\sqrt{2}$ is not of degree $2$ (it would have to be expressible in the form $a + b\sqrt{d}$ for some squarefree integer $d$, with $a,b\in \mathbb{Q}$, and this is impossible. So you are certainly right that it is degree $4$.

Now, here your method is not actually "guessing", but "working." Suppose $f(x)$ is a the minimal polynomial of $i+\sqrt{2}$, and consider the splitting field of $f(x)$, $K$. Complex conjugation gives an automorphism of $K$, and maps one root of $f(x)$ to another root of $f(x)$. That means that $-i+\sqrt{2}$ must also be a root of $f(x)$.

Likewise, the automorphism of $\mathbb{Q}(\sqrt{2})$ that fixes $\mathbb{Q}$ and maps $\sqrt{2}$ to $-\sqrt{2}$ extends to an automorphism of $K$ which fixes $f(x)$, and maps any root of $f(x)$ to a root of $f(x)$. So both $i-\sqrt{2}$ (the image of $i+\sqrt{2}$) and $-i-\sqrt{2}$ (the image of $-i+\sqrt{2}$) must be roots of $f(x)$.

This gives you four roots of $f(x)$, which you know to be of degree $4$, so that gives the four roots and hence $f(x)$.

For more general comments, see this previous question. You can consider the powers of $i+\sqrt{2}$ until you get that $1, \alpha,\alpha^2,\ldots,\alpha^n$ is linearly dependent over $\mathbb{Q}$, and use the linear dependency to get the polynomial.

Added. In this case, we have: $$\begin{align*} i+\sqrt{2} &= i+\sqrt{2}\\ (i+\sqrt{2})^2 &= 1 + 2i\sqrt{2}\\ (i+\sqrt{2})^3 &= 5i - \sqrt{2}\\ (i+\sqrt{2})^4 &= -7 + 4i\sqrt{2}\\ &= -9 + 2(1+2i\sqrt{2}) = -9 + 2(i+\sqrt{2})^2. \end{align*}$$ This means that $\alpha=i+\sqrt{2}$ satisfies $\alpha^4 = -9+2\alpha^2$, or that $\alpha$ is a root of $f(x)=x^4 -2x^2 + 9$.

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I don't see the link to the previous question. – Ross Millikan Mar 25 '11 at 14:08
@Ross This is done. – Did Mar 25 '11 at 14:13
@Ross: Sorry; had to go teach... – Arturo Magidin Mar 25 '11 at 14:54

There are various methods, e.g. using automorphisms, taking the characteristic polynomial of $\ z \mapsto (i+\sqrt{2})\ z,\,$ undetermined coefficients, etc. Here the simplest may be repeated squaring: $ \ x - i = \sqrt{2}\, $ $\,\overset{(\ \ )^2}\Rightarrow$ $ x^2 - 2\ i\ x -1 = 2\ $ or $ \ x^2 - 3 = 2\ i\ x,\,$ which, squared, yields the result.

Uupdate $\ $ Since the terseness of the "repeated squaring" method seems to have possibly confused at least one reader, perhaps it may help to elaborate a bit.

Theorem $\ $ Suppose that $ \,R\,$ is a ring which contains two elements $ \, w,\, i\, $ that satisfy $ \, w^2 = 2,\ \ i^2 = -1\,.\ $ Then $\ x = w + i\ $ satisfies $\ x^4 + 9 = 2x^2\,.$

Proof $\ $ Squaring $ \ x - i = w\ $ yields $ \ x^2 - 2\,i\ x - 1\, =\, 2,\,$ or $ \ x^2 - 3\ =\ 2\,i\, x.\,$ Squaring this yields $ \ x^4 -6\, x^2 + 9\ =\, -4\, x^2,\ $ so $ \ x^4 + 9\, =\, 2\, x^2\,.\quad$ QED

Remark $\ $ Notice that the above theorem may apply in rings other than $ \, \mathbb Q(i,\sqrt{2}).\, $ For example, $\,$ in $ \, \mathbb Z/17 =\,$ integers mod $17\,$ we can choose $ \ w = 6,\ i = -4\ $ and conclude that $ \ x = w+i = 2\, \Rightarrow\, x^4 -2\ x^2 + 9\, =\, 0.\ $ Indeed $ \ 16 - 8 + 9\, =\, 17 \equiv\ 0\pmod{ 17}.$

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@Bill You use $\mathtt{x}$ for two different purposes. And you might wish to correct the repeated squaring. – Did Mar 25 '11 at 14:11
@Didier: Your comment makes no sense to me. – Bill Dubuque Mar 25 '11 at 14:15
@Bill The symbols $\mathtt{x}$ did not appear as factors of $2\mathrm{i}$ when I first looked at your comment. Now they do, so the second part of my comment is moot. Re the first part, you use $\mathtt{x}$, first as the running argument of the transformation $\mathtt{x}\to(\mathrm{i}+\sqrt{2})\mathtt{x}$, then as $\mathtt{x}=\mathrm{i}+\sqrt{2}$. – Did Mar 25 '11 at 14:27
@Didier: I still see no problems. – Bill Dubuque Mar 25 '11 at 14:34
@Bill You did not define the symbol $\mathbb{x}$ in the sense you use it in the last sentence of your post. Since you used the same symbol for something completely different in the sentence just before, this might be seen as an unfortunate choice of notations. (Sorry if my first comments on this were too cryptic.) – Did Mar 25 '11 at 14:41

The way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\bar{z}) = \bar{P(z)}$ for any complex number $z$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $-i + \sqrt{2}$. Similarly, suppose $z = a + b\sqrt{2}$, with $a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\sqrt{2}) = c + d\sqrt{2}$, one has $P(a - b\sqrt{2}) = c - d\sqrt{2}$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $i - \sqrt{2}$, and if $-i + \sqrt{2}$ is a root of $P(z)$, so is $-i - \sqrt{2}$.

The upshot is that if $i + \sqrt{2}$ is a root of $P(z)$, so are $-i + \sqrt{2}$, $i - \sqrt{2}$, and $-i - \sqrt{2}$. Thus the minimal polynomial of $i + \sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.

In some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.

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