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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!

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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!

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Dear Amitesh, you missed a $ sign, for \mathbb{Q}. I have edited it. Hope that isn't a concern –  user9413 Jul 8 '11 at 12:11
    
@Chandru: Dear Chandru, absolutely not. Thank you very much for your edit! –  Amitesh Datta Jul 8 '11 at 12:12
    
@Amitesh Datta Hi, this is sort of a side comment really, but is your mentor Elizabeth Ormerod? Just wandering, she mentioned of you to me once when I met her. It would be interesting to meet you at university one day (Sorry I had to say this here). –  user38268 Jul 8 '11 at 12:56
    
@D Lim: Actually, no but I do know her well (she supervised me for an ASC on finite group theory last semester). My mentor is James Borger. Anyway, I would be happy to meet you one day; unfortunately, I am not in Canberra at the moment. I will be in Canberra on July 26th. You could send me an email; my email address is amiteshdatta@gmail.com. –  Amitesh Datta Jul 8 '11 at 13:03
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@confused Thanks for the kind words! Yes, a polynomial "over $\mathbb{Q}$" is a polynomial with coefficients in $\mathbb{Q}$. Similarly, a complex number $x$ is algebraic "over $\mathbb{Q}$" if there is a polynomial $p$ with coefficients in $\mathbb{Q}$ such that $p(x)=0$ etc. –  Amitesh Datta Jul 8 '11 at 13:53
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We show how to exploit symmetries to produce an answer. Although we only deal with the specific question, we deal with it in a structured way that at least hints at generalization.

Note that $\sqrt{2}+\sqrt{3}$ is the solution of $x-(\sqrt{2}+\sqrt{3})=0$. But $\sqrt{2}$ and $-\sqrt{2}$ are friends and like to travel in pairs, as do $\sqrt{3}$ and $-\sqrt{3}$.

In acknowledgement of this, let $$P(x)=(x-(\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3}))(x-(\sqrt{2}-\sqrt{3})).$$

Intuitively speaking at least, $P(x)$ should have integer coefficients. Why? First observe that the coefficients of a polynomial are symmetric functions of the roots.

Now imagine "simplifying" $P(x)$. Each coefficient of $P(x)$ must be of the shape $$A+B\sqrt{2}+ C\sqrt{3} + D\sqrt{2}\sqrt{3},$$ where $A$, $B$, $C$, and $D$ are integers. But this expression must remain unchanged if we replace $\sqrt{2}$ by $-\sqrt{2}$ and/or $\sqrt{3}$ by $-\sqrt{3}$, so we should have $B=C=D=0$. Thus every coefficient of $P(x)$ should be an integer.

Until we back up this kind of intuition with general theorems, our discussion has a speculative character. However, in this case, we can easily multiply out $P(x)$ and check directly that it has integer coefficients. The product of the first two terms is $x^2-2\sqrt{3}\,x+1$, and the product of the next two terms is $x^2+2\sqrt{3}\,x +1$. Thus $$P(x)=x^4-10x^2+1,$$ and our intuition has been fully verified.

Comment: All the speculative elements above can be put on a firm foundation. With the proper theoretical background in place, we would not have needed to compute at all. "Symmetries" play a central role in the study of algebraic numbers. This was a key insight of Lagrange, which led ultimately to the breakthroughs by Abel and Galois.

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Call $x= \sqrt{2} + \sqrt{3}$, then by your equation we have $$x^{2} = 5 +\sqrt{24} \Longrightarrow (x^{2}-5)^{2}=24$$

For a proof of sum of $2$ algebraic numbers being algebraic, please see:

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You should just proceed until you obtain $(x^2-5)^2 = 24$.

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Note $\rm\ \alpha = \sqrt{2}\:+\sqrt{3}\in F = \mathbb Q(\sqrt{2},\sqrt{3})\:.\:$ But $\rm\:F = \mathbb Q\langle1,\sqrt{2},\sqrt{3},\sqrt{6}\rangle\:$ is a $\rm\:\mathbb Q$-vector space of dimension $4\:,\:$ so the $\:5\:$ elts $\:1,\alpha,\alpha^2,\alpha^3,\alpha^4$ are $\rm\:\mathbb Q$-linearly dependent. Thus $\rm\:f(\alpha) = 0\:$ for some $\rm\:0 \ne f(x)\in\mathbb Q[x]\:$ of degree $\le 4\:.\:$ This proof by dimension works generally - see Zhen's answer.

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My answer (above) also discusses the method of "proof by dimension" (as an exercise) ... –  Amitesh Datta Jul 9 '11 at 3:44
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It is true that the sum and product of two algebraic numbers is algebraic. Thus, the algebraic numbers form a field.

The easiest way to see this is, however, indirect. I will assume you know some linear algebra over fields other than $\mathbb{R}$. Let $\mathbb{Q}$ be the field of rational numbers, and for simplicity we shall assume that $\mathbb{Q} \subset \mathbb{C}$. Let $\alpha \in \mathbb{C}$. By $\mathbb{Q}[\alpha]$ we mean the smallest subring of $\mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It should be fairly obvious that $\mathbb{Q}[\alpha]$ consists of all finite sums of the form $c_0 + c_1 \alpha + \cdots + c_r \alpha^r$, for all $r \in \mathbb{N}$, and so we see that $\mathbb{Q}[\alpha]$ is a vector space over $\mathbb{Q}$, and $\{ 1, \alpha, \alpha^2, \ldots \}$ is a spanning set. Now, suppose $\alpha$ is algebraic. Then $\alpha$ satisfies a polynomial, so in fact there is a finite number $n$ so that $\{ 1, \alpha, \ldots, \alpha^{n-1} \}$ is a basis for $\mathbb{Q}[\alpha]$. Conversely, if $\mathbb{Q}[\alpha]$ is finite-dimensional, then any $x$ in $\mathbb{Q}[\alpha]$ must be algebraic: since, for $n$ large enough, $\{ 1, x, \ldots, x^n \}$ will not be linearly independent, and we will be able to obtain a non-trivial equation which relates the powers of $x$, thus a witness for the algebraicity of $x$. So we have connected algebraicity with finite-dimensionality.

Now, let $\alpha$ and $\beta$ be two algebraic numbers. By $\mathbb{Q}[\alpha, \beta]$ we mean the smallest subring of $\mathbb{C}$ containing $\mathbb{Q}$ and $\alpha$ and $\beta$. But any such ring must contain $\alpha + \beta$ and $\alpha \beta$, so if we can show $\mathbb{Q}[\alpha, \beta]$ is finite-dimensional then we are done. But it has to be: we know any element of $\mathbb{Q}[\alpha]$ can be written as a finite sum of powers of $\alpha$ and similarly for $\beta$, thus, if $\{ 1, \alpha, \ldots, \alpha^{n-1} \}$ is a basis for $\mathbb{Q}[\alpha]$ and $\{ 1, \beta, \ldots, \beta^{m-1} \}$ is a basis for $\mathbb{Q}[\beta]$, $\{ \alpha^a \beta^b : 0 \le a < n, 0 \le b < m \}$ must be a spanning set for $\mathbb{Q}[\alpha, \beta]$. Thus the dimension of $\mathbb{Q}[\alpha, \beta]$ is at most the product of the dimensions of $\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\beta]$, and in particular $\alpha + \beta$ and $\alpha \beta$ are algebraic.

The only thing left to do is to show that $1 / \alpha$ is algebraic if $\alpha$ is algebraic and non-zero, but this is straightforward enough: given a polynomial $\alpha$ satisfies, we can manipulate it to find a polynomial $1 / \alpha$ satisfies. (Exercise for the reader!)

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There is another proof that every non-zero element in $\mathbb{Q}(a)$ is invertible and it too uses linear algebra: Consider the map $x \mapsto ax$ on $\mathbb{Q}(a)$. This map is linear and has trivial kernel, since otherwise $1,a,\dotsc,a^{n-1}$ would be linearly dependant—contradicting that the degree of the minimal polynomial of $a$ is $n$. We conclude that $a$ has an inverse in $\mathbb{Q}(a)$ and the invertability of all other non-zero elements follows. You can read a bit more detail at this page: dpmms.cam.ac.uk/~wtg10/galois.html –  kahen Jul 8 '11 at 20:17
    
PS: Isn't it customary to write "$K(a)$ for $K$ with $a$ adjoined and $K[a]$ for the ring of polynomials in one variable over $K$? –  kahen Jul 8 '11 at 20:17
    
@kahen: $K(a)$ is the smallest field containing $K$ and $a$, and I didn't want to muddy things. –  Zhen Lin Jul 8 '11 at 23:38
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