In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. For example, I tried $(\sqrt{2}+\sqrt{3})^{2}=5+\sqrt{24}$ and then I tried to think of a polynomial of form $ax^2-bx^{0}=0$ e.g. $c(\sqrt{2}+\sqrt{3})^{2}-c(5+\sqrt{24})=0$ that would work, but couldn't find the $c$ value that would make $b=c(5+\sqrt{24})$ integer, but couldn't find one. Maybe I can somehow conjecture that the sum of two algebraic numbers must be algebraic, too, but I was wondering if there's a way to find a polynomial to show this. Thanks!
Let $x=\sqrt{2}+\sqrt{3}$. Note that $x^2=2+2\sqrt{6}+3$ and therefore $x^2-5=2\sqrt{6}$. We can square both sides of this equation and obtain $(x^2-5)^2=24$. You should now be able to show that $x$ is an algebraic number (over $\mathbb{Q}$). (In fact, it is instructive to expand this equation and obtain a monic polynomial $p$ of degree $4$ such that $p(x)=0$.)

The following exercises are relevant:

Example-Based Exercises:

Exercise 1: Prove that the following numbers are algebraic (over $\mathbb{Q}$):

(a) $\sqrt{2}+\sqrt{5}$

(b) $\sqrt{2}+\sqrt{4}$

(c) $\sqrt{2}+\sqrt[3]{3}$

The degree of an algebraic number $x$ is defined to be the least positive integer $n$ such that there is a monic polynomial $p$ of degree $n$ with $p(x)=0$. Compute the degrees of each of the algebraic numbers above (in (a), (b) and (c)). Prove that your answer is correct in each case.

Exercise 2: Let $p_1,\dots,p_n$ be distinct (positive) prime numbers. Prove that the number $x=\sqrt{p_1}+\cdots+\sqrt{p_n}$ is algebraic. What is the degree of $x$?

Exercise 3: Let $x$ be an algebraic number and let $q$ be a rational number. Prove that the degree of $x$ equals the degree of $x+q$. 

Exercise 4: Let $x=\cos(\frac{2\pi}{n})$ for some positive integer $n$. Is $x$ algebraic (over $\mathbb{Q}$)? If not, prove it, and if so find the degree of $x$. (Your answer will depend on $n$.)

Theoretical Exercises:

Exercise 5: Let $x\in\mathbb{C}$ and define $\mathbb{Q}[x]=\{f(x):f\text{ is a polynomial with rational coefficients}\}$.

(a) Prove that $\mathbb{Q}[x]$ is a finite dimensional vector space over $\mathbb{Q}$ if and only if $x$ is algebraic over $\mathbb{Q}$. (Hint (for one direction): if $x$ is algebraic over $\mathbb{Q}$ and if $a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n=0$ where $a_i\in\mathbb{Q}$ for all $0\leq i\leq n$, show that $\{1,x,x^2,\dots,x^{n-1}\}$ spans $\mathbb{Q}[x]$ as a $\mathbb{Q}$-vector space.)

(b) Prove that $\mathbb{Q}[x]$ is a field if $x$ is algebraic over $\mathbb{Q}$. (Hint: it suffices to prove that $x$ has a multiplicative inverse. Let $p$ be a monic polynomial of minimal degree (with coefficients in $\mathbb{Q}$) such that $p(x)=0$. Prove that $p(0)\neq 0$ if $x\neq 0$.)

(c) Prove that if $\mathbb{Q}\subseteq K\subseteq L$ is a tower of fields and if $K$ is a finite dimensional $\mathbb{Q}$-vector space, $L$ is a finite dimensional $K$-vector space, then $L$ is a finite dimensional $\mathbb{Q}$-vector space.

(d) Prove that the sum and product of two algebraic numbers (over $\mathbb{Q}$) is algebraic. (Hint: if $x$ and $y$ are algebraic, use (c) to show that $\mathbb{Q}[x,y]=\{f(x,y):f\text{ is a polynomial in two variables with rational coefficients}\}$ is a finite dimensional $\mathbb{Q}$-vector space. (a) is relevant.)

Challenge: If $x$ and $y$ are algebraic numbers (over $\mathbb{Q}$), can you find an explicit polynomial $p$ with rational coefficients such that $p(x+y)=0$? Can you find an explicit polynomial $q$ with rational coefficients such that $q(xy)=0$?

I hope this helps!
