TITLE: Prove: if $a$ and $b$ are algebraic, then $a + b$, $a - b$ and ab are also algebraic
QUESTION [5 upvotes]: I have to prove the following:
If $a, b \in \mathbb{C}$ and are both algebraic over $\mathbb{Z}$, then:

$a + b$ is algebraic over $\mathbb{Z}$ 
$a - b$ is algebraic over $\mathbb{Z}$
$ab$ is algebraic over $\mathbb{Z}$

I tried this for the first one: 
$a, b$ are algebraic, so there is a $f(x) \in \mathbb{Z}[x]$ with $f(a) = 0$, and also a $g(x) \in \mathbb{Z}[x]$ with $g(b) = 0$.
I don't know how to prove these three statements. It would be very helpful if you could prove one of them for me in a simple way, and then I will be able to do the rest of them myself.
Thanks in advance!

REPLY [2 votes]: Probably the easiest way to tackle this is by saying:
$a$ is algebraic implies that the field extension $\mathbb{Q}(a):\mathbb{Q}$ is algebraic, and thus $[\mathbb{Q}(a):\mathbb{Q}] < \infty$.
Similarly, we have $[\mathbb{Q}(b):\mathbb{Q}] < \infty$.
So by the tower law, $[\mathbb{Q}(a,b):\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}] < \infty.$ 
This implies $\mathbb{Q}(a,b):\mathbb{Q}(a)$ is an algebraic field extension, so all terms of the form $r + sa +qb$, $r,s,q \in \mathbb{Q}$ are algebraic.