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.

  • $\begingroup$ I don't think there is a "simple" way. To show that a number is algebraic, you need to show that it is the root of a polynomial. The only facts given are that the two numbers you add are also the roots of a polynomial. But finding a polynomial whose root is the sum of the roots of two other polynomials is a hard problem. Which is why algebra textbooks usually don't try to construct them, and instead argue indirectly via the degree of certain field extensions. But for that you need at least some basic abstract algebra (and I mean basic, like the first chapter of a textbook). $\endgroup$ – Vercassivelaunos Sep 25 at 8:56

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 $0\le k<n$.

| cite | improve this answer | |
  • $\begingroup$ This is what I've been looking for. $\endgroup$ – Sphere Sep 25 at 13:59
  • $\begingroup$ @Sphere It's really the same as Patrick da Silva's argument at math.stackexchange.com/questions/155122/… but just written in a more naive manner. $\endgroup$ – Angina Seng Sep 25 at 14:06
  • $\begingroup$ But I don't know tensor product. Anyway I understood your proof. $\endgroup$ – Sphere Sep 25 at 14:09

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 $\alpha+\beta$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.