# Find a basis for $\mathbb Q(i \sqrt{2})$ over $\mathbb Q$ and describe its elements.

Find a basis for $\mathbb Q(i \sqrt{2})$ over $\mathbb Q$ and describe its elements.

So I need to find a minimum polynomial of $i\sqrt{s}$ over $\mathbb Q$ which I believe has degree 2 (I don't know why or how to find it I'm lost on this topic). Then I need to find a linear combination using the minimum polynomial and that will be the elements of the basis. I know for $\mathbb Q\sqrt{2}$ that the elements are of the form $a+b\sqrt{2}$. Would $\mathbb Q (i\sqrt{2})$ consist of elements of the form $a+bi\sqrt{2}$?

• Yes, the elements would be of the form $a + bi\sqrt2$. However, we need conclusive proof that that covers all of it. How do we know that it's not $a + bi\sqrt2 + c(i\sqrt2)^2+d(i\sqrt2)^3$ or something? At this point it might be a good idea to look for the minimal polynomial of $i\sqrt2$ over $\Bbb Q$, or just any polynomial. Can you think of a polynomial $f(x)$ with rational coefficients such that $f(i\sqrt2) = 0$? – Arthur Nov 27 '17 at 14:26
• @Arthur I'm not sure this is the right approach but would we take $a=i\sqrt{2}\rightarrow a^2=-1\cdot 2 \rightarrow a^2+2=0$ so $x^2+2$ is a polynomial? – K Math Nov 27 '17 at 14:28
• Exactly! There is some minor fiddling to show that what you have there is the minimal polynomial (which is the same as showing that that polynomial doesn't factor into smaller degree polynomials with rational coefficients), but otherwise that's it. – Arthur Nov 27 '17 at 14:31
• The only forms that are even possible are like I wrote earlier, something like $a + bi\sqrt2 + c(i\sqrt2)^2+d(i\sqrt2)^3$ (possibly more terms). However, the minimal polynomial being $x^2 + 2$ is saying that $(i\sqrt2)^2 + 2 = 0$ or $(i\sqrt2)^2 = -2$, so you can rewrite $c(i\sqrt2)^2+d(i\sqrt2)^3$ into $-2c-2di\sqrt2$. In other words, any term of order $2$ or higher can have its degree decreased. Finally, you're left with just something of the form $a+bi\sqrt2$. Also, the minimal polynomial being degree $2$ means that you can't do anything more with these lower-order terms. – Arthur Nov 27 '17 at 14:41
• For a second example, consider an element $p$ with minimal polynomial over $\Bbb Q$ of order $5$. The field $\Bbb Q(p)$ has elements of the form $a+bp+cp^2+dp^3+ep^4$. Any representation with higher order terms may be reduced by using the minimal polynomial (the polynomial may be rewritten to $p^5 = \cdots$). And there is no more reduction or alternate coefficients possible, because that would imply a relation between $1, p, p^2, p^3, p^4$, which would give rise to a polynomial of degree $4$ or less, which can't happen because the minimal polynomial has degree $5$. – Arthur Nov 27 '17 at 14:43

In order to make it immediately obvious that the extension is degree 2, it helps to write $\Bbb{Q}(\sqrt{-2})$ instead of $\Bbb{Q}(i\sqrt{2})$.