elementary proof that sum of algebraic numbers is algebraic(Real Analysis and Foundations) How can one prove that the sum of two algebraic numbers is an algebraic number without using theories of algebra? I don't know about (abstract) algebra. Actually, this is an exercise of an analysis book(Real Analysis and Foundations, 4th edition, Steven G.Krantz, p67 exercise 6), so there must be a way of solving this in a relatively simple manner.
(Proof Verification) Prove that the sum of two algebraic numbers is algebraic. and Elementary proof for the theorem that field of algebraic numbers is closed use theories of algebra, How to prove that the sum and product of two algebraic numbers is algebraic? uses algebra and tensor product(which I don't know).
What I know is calculus, linear algebra, and a little set theory and analysis.
 A: I'll do an example, with I hope is general enough to demonstrate the general procedure.
Let $\alpha$ have minimum polynomial $x^2+x+1$ and $\beta$ have minimal polynomial $x^3-x-1$. I'll construct a matrix $M$ with integer entries having $\alpha+\beta$ as an eigenvalue.
This will show that $\alpha+\beta$ is a zero of the characteristic equation of $M$,
and as that has integer coefficients, meaning that $\alpha+\beta$ is an algebraic integer.
I will construct $M$ such that
$$M\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
=(\alpha+\beta)\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
.$$
Then $\alpha+\beta$ will be an eigenvalue of $M$.
Using $\alpha^2=-\alpha-1$ and $\beta^3=\beta+1$ we can express all entries
in the vector in the right as integer linear combinations of the purported eigenvector.
In detail,
$$(\alpha+\beta)1=\alpha+\beta,$$
$$(\alpha+\beta)\alpha=-1-\alpha+\alpha\beta,$$
$$(\alpha+\beta)\beta=\alpha\beta+\beta^2,$$
$$(\alpha+\beta)\alpha\beta=-\beta-\alpha\beta+\alpha\beta^2,$$
$$(\alpha+\beta)\beta^2=1+\beta+\alpha\beta^2,$$
$$(\alpha+\beta)\alpha\beta^2=\alpha+\alpha\beta-\beta^2-\alpha\beta^2$$
Then (E&OE)
$$M=\pmatrix{0&1&1&0&0&0\\
-1&-1&0&1&0&0\\
0&0&0&1&1&0\\
0&0&-1&-1&0&1\\
1&0&0&0&0&1\\
0&1&0&1&-1&-1}$$
and $\alpha+\beta$ is an eigenvalue of $M$.
As an exercise, find $N$ with integer entries and
$$N\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
=\alpha\beta\pmatrix{1\\\alpha\\\beta\\\alpha\beta\\\beta^2\\\alpha\beta^2}
.$$
In general, with $\alpha$ and $\beta$ having degrees $m$ and $n$,
take the vector whose entries are $\alpha^j\beta^k$ with $0\le j<m$
and $0\le k<n$.
A: You can have a simple proof  with linear algebra: a real number $\alpha$ is algebraic if and  only if  $\mathbf Q[\alpha]$ is a finite dimensional $\mathbf Q$-vector space.
Now, if $\alpha$ and $\beta$ are algebraic, $\mathbf Q[\alpha]$ and $\mathbf Q[\beta]$ are finite dimensional over $\mathbf Q$. You easily deduce that $\:\mathbf Q[\alpha,\beta]=\mathbf Q[\alpha][\beta]\:$ is finite dimensional over $\mathbf Q[\alpha]$. As
$$\dim_{\mathbf Q}\bigl(\mathbf Q[\alpha,\beta]\bigr) =\dim_{\mathbf Q}\bigl(\mathbf Q[\alpha]\bigr)\cdot\dim_{\mathbf Q[\alpha]}\bigl(\mathbf Q[\alpha,\beta]\bigr)<\infty,$$
we see that all elements of $\mathbf Q[\alpha,\beta]$ are algebraic numbers, in particular $\alpha+\beta$.
