# Prove that $\mathbb Q(3+i) = \mathbb Q(1-i)$.

Dealing with field extensions, in which $\mathbb Q(u)$ is the extension of the rationals in which $u$ is a root.

-

One other approach: Note that $\mathbb{Q}(u)$ contains all of $\mathbb{Q}$ and $u$, so $u + q$ is in the extension as well, for any $q\in \mathbb{Q}$. Hence, $$\mathbb{Q}(3+i) \supset \mathbb{Q}(i) \subset \mathbb{Q}(1-i).$$

But each of these fields is a degree 2 extension (can you show why?), so...

If you don't know how to work with degrees easily, you can use a similar method getting the first inclusions, to show the other direction of the inclusions.

Edit: The extensions are degree two for both of those reasons actually! One way of defining the field extension $\mathbb{Q}(u)$ is to use a basis (since every extension is a vector space over the base field) . In this case, each field has the respective basis $$\mathbb{Q}(3+i) : \{1,3+i\},$$ $$\mathbb{Q}(i) : \{1,i\},$$ $$\mathbb{Q}(1-i) : \{1,1-i\}.$$ But these bases are all the "same", up to a minor change. I.e., for any element $q = a + b(3 + i) \in \mathbb{Q}(3+i)$, we have $q = (a + 3b) + bi \in \mathbb{Q}(i)$, etc...

More generally, the extension $\mathbb{Q}(u)$ can be defined as the (field) vector space with basis $\{1,u,u^2,\ldots\}$, and in each of the three cases above, $u^2,u^3,\ldots$ are linear combinations of $1$ and $u$, so our basis for $\mathbb{Q}(u)$ is simply $\{1,u\}$.

On the other hand, each of the elements we are adjoining satisfies a monic irreducible degree two polynomial, though be careful, as the irreducibility of the polynomial is very important here! For instance, for $\mathbb{Q}(i)$ we have $$f(i)(i^2 + 1) = 0$$ for any polynomial $f(x) \in \mathbb{Q}[x]$, and so the only way we can say $i$ has degree 2 (and not all degree greater than or equal to 2), is to require the irreducibility of the polynomial. The monic requirement isn't as important, but it's used so that there is only one (monic) irreducible polynomial to consider for any algebraic element we adjoin, called the minimal polynomial of that element.

One can then show that both of these definitions of degree (i.e., number of generators for the vector space, or degree of the minimal polynomial) are equivalent.

I don't know what book you're using, though I've had good experiences with (undergrad) Gallian's Contemporary Abstract Algebra and (grad) Abstract Algebra.

If these books are not available to you, I haven't used it, but other books by Milne have worked well for me, so I can suggest his Field and Galois theory notes in hopes they work well for you. As a bonus they're available on his webpage here. Also, I'm sure there are lots of expository that may cover bits here and there if you need a little help (such as Keith Conrad's notes here). Of course, after all these references, this site is specifically intended to help people when they're stuck, so this is definitely a good place as well, so don't feel I'm trying to get you to go somewhere else! As per your comment, I'm sure everyone with an account of this has made "embarassing" slip ups at least once.. (I sure have!) so don't worry :)

Hope this helps!

-
Thank you! This was embarrassingly easy in hindsight, but I just kind of got thrown into this lesson and the book I'm using is extremely unclear. Out of curiosity, is it a degree two extension because every number can be expressed with factors of {1,i} which has two elements? Or is it because i is the solution to x^2 + 1 = 0, and that's a degree two polynomial? Sorry if I'm compounding my earlier question with another dumb question, but I'm pretty much flying blind here. –  Eyeball McCool Nov 16 '12 at 5:31
@EyeballMcCool, I've added more to the answer, as the comment boxes are far too small! :) –  Alex Nov 16 '12 at 6:39
One half of the job: Since $4$ is in $\mathbb{Q}$, it follows that $4-(3+i)$ is in $\mathbb{Q}(3+i)$. So $1-i \in \mathbb{Q}(3+i)$, and therefore $\mathbb{Q}(1-i)\subseteq \mathbb{Q}(3+i)$.