Prove: if $a$ and $b$ are algebraic, then $a + b$, $a - b$ and ab are also algebraic

Question

I have to prove the following:

If $a, b \in \mathbb{C}$ and are both algebraic over $\mathbb{Z}$, then:

1. $a + b$ is algebraic over $\mathbb{Z}$

2. $a - b$ is algebraic over $\mathbb{Z}$

3. $ab$ 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!

Answer 1

The easiest way is by using these facts, which are easily proved:

• $a\in\mathbb C$ is algebraic iff $\mathbb Q[a]$ is finite-dimensional over $\mathbb Q$.

• If $a$ and $b$ are algebraic then $\mathbb Q[a,b]$ is finite-dimensional over $\mathbb Q$.

The result then follows because $a\pm b$ and $ab$ are in $\mathbb Q[a,b]$ and subspaces of finite-dimensional spaces are finite-dimensional themselves.

Answer 2

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.