What is a general way to find out whether a number obtained by a finite combination of algebraic operations is algebraic?

Question

What is a general way to get a integers inside a radical with + or - operation(the numbers adding or subtracting each others, for example, $\sqrt5 +\sqrt7$ is this type of numbers)allow is algebraic or not? In another word, prove any number obtained by a finite combination of algebraic operations (addition, multiplication, root extraction) is algebraic

Answer 1

1. Every $r\in\mathbb{Q}$ is algebraic over $\mathbb{Q}$.

   Proof. $r$ is a root of $x-r$, a polynomial with rational coefficients.

2. If $a$ is algebraic over $\mathbb{Q}$ and $n$ is a positive integer, then $\sqrt[n]{a}$ is algebraic over $\mathbb{Q}$ (where $\sqrt[n]{a}$ is any fixed complex $n$th root of $a$).

   Proof. Since $a$ is algebraic, there is a polynomial $f(x) = x^m+a_{m-1}x^{m-1}+\cdots+ a_0$, with $a_i\in\mathbb{Q}$, wuch that $f(a)=0$. Then $\sqrt[n]{a}$ is a root of $$x^{mn} + a_{m-1}x^{(m-1)n} + \cdots + a_1x^n + a_0,$$ hence $\sqrt[n]{a}$ is algebraic.

3. If $a$ is algebraic over $\mathbb{Q}$ and $k$ is in $\mathbb{Q}$, then $ka$ is algebraic over $\mathbb{Q}$.

   Proof. If $k=0$, there is nothing to do. If $k\neq 0$, and $f(x)=x^m+a_{m-1}x^{m-1}+\cdots + a_0$ has $a$ as a root, then $$\frac{1}{k^m}x^m + \frac{a_{m-1}}{k^{m-1}}x^{m-1}+\cdots + \frac{a_1}{k}x + a_0$$ has $ka$ as a root.

4. If $a$ and $b$ are algebraic over $\mathbb{Q}$, then $a+b$ is algebraic over $\mathbb{Q}$.

   If $f(x)$ and $g(x)$ are any two polynomials, then their resultant $R(f,g)$ is the product of all $a-b$ with $a$ a root of $f$ and $b$ a root of $g$. Suitably interpreted, you can use it to obtain a polynomial whose roots are precisely the elements of the form $a-b$ with $a$ a root of $f$ and $b$ a root of $g$. The resultant can be computed via the determinant Sylvester matrix, and in particular if $f$ and $g$ have integer (or rational) coefficients, then so does the polynomial you obtain from $R(f,g)$. $R(f,g)=0$ if and only if $f$ and $g$ have a zero in common.

   Then computing the resultant of $f(x)$ and $g(z-x)$ gives a polynomial (in $z$) with rational coefficients that has, among its roots, the root $a+b$; hence, $a+b$ is algebraic.

5. If $a$ is algebraic and $a\neq 0$, then $\frac{1}{a}$ is algebraic

   Proof. If $f(x) = x^m + a_{m-1}x^{m-1}+\cdots + a_1x + a_0$ is a polynomial with rational coefficients and $f(a)=0$, then $$x^mf\left(\frac{1}{x}\right) = 1 + a_{m-1}x + \cdots + a_1x^{m-1} + a_0x^m$$ is a polynomial with rational coefficients that has $\frac{1}{a}$ as a root.

6. If $a$ and $b$ are algebraic over $\mathbb{Q}$, then $ab$ is algebraic over $\mathbb{Q}$.

   Proof. If $f(x)$ has rational coefficients and $a$ as a root, and $g(x)$ has rational coefficients and $b$ as a root, then the resultant of $f(x)$ and $x^{\deg(g)}g(\frac{t}{x})$ is a polynomial in $t$ with rational coefficients that has $xy$ as a root. Hence $xy$ is algebraic.

Thus, every complex number obtained as the result of doing a finite combination of addition, multiplication, and root extraction of algebraic numbers is algebraic. You'll note that the above gives you a method for explicitly producing a polynomial that can "witness" the algebraicity of the result, though you may not want to carry it out in practice.

(Note: The above easily generalizes if we replace $\mathbb{Q}$ with an arbitrary field, and $\mathbb{C}$ with an arbitrary extension: if $F$ is a field, and $F\subseteq K$, then the collection of all elements of $K$ that are algebraic over $F$ forms a field that contains $F$.)

For example: $\sqrt{3}+\sqrt{5}$; a polynomial with $\sqrt{3}$ as a root is $f(x)=x^2-3$. A polynomial with $\sqrt{5}$ as a root is $g(x)=x^2-5$. We take the resultant of $f(x)$ and $g(z-x) = (z-x)^2 - 5 = x^2 - 2zx + (z^2-5)$: $$\begin{align*} R(f(x),g(z-x)) &= \left|\begin{array}{crcc} 1 & 0 & -3 & 0\\ 0 & 1 & 0 & -3\\ 1 & -2z & z^2-5 & 0\\ 0 & 1 & -2z &z^2-5 \end{array}\right|\\ &\strut\\ &= \left|\begin{array}{rcc} 1 & 0 & -3\\ -2z & z^2-5 & 0\\ 1 & -2z & z^2-5 \end{array}\right| + \left|\begin{array}{ccc} 0 & -3 & 0\\ 1 & 0 & -3\\ 1 & -2z & z^2-5 \end{array}\right|\\ &\strut\\ &= \left|\begin{array}{cc} z^2-5 & 0\\ -2z & z^2-5 \end{array}\right| -3\left|\begin{array}{rc} -2z & z^2-5\\ 1 & -2z \end{array}\right| +3\left|\begin{array}{cc} 1 & -3\\ 1 & z^2-5 \end{array}\right|\\ &\strut\\ &= (z^2-5)^2 -3(4z^2 - z^2+5) +3(z^2-5+3)\\ &\strut\\ &= z^4 - 10z^2 + 25 - 9z^2 - 15 + 3z^2 -6\\ &\strut\\ &= z^4 - 16z^2 + 4. \end{align*}$$

Indeed, $\sqrt{3}+\sqrt{5}$ satisfies this polynomial: you can do it directly by substitution, or letting $a=\sqrt{3}+\sqrt{5}$, note that $a^2 = 8+2\sqrt{15}$, hence $a^2-8 = 2\sqrt{15}$, so $(a^2-8)^2 = 60$. Thus, $a^4 - 16a^2 + 64=60$, so $a$ satisfies $x^4 - 16x^2 + 4$, the same polynomial we found above.

The polynomial we get need not be the minimal polynomial: again with $\sqrt{3}$ and $\sqrt{5}$, the minimal polynomial of $\sqrt{3}\sqrt{5}$ is of course $x^2-15$. Using the resultant method, we need to compute the resultant of $f(x) = x^2-3$ and of $$x^2g\left(\frac{t}{x}\right) = x^2\left(\frac{t^2}{x^2} - 5\right) = t^2-5x^2.$$ We obtain: $$\begin{align*} \mathrm{Res}\left(f(x),x^2g\left(\frac{t}{x}\right)\right) &= \left|\begin{array}{rrrr} 1 & 0 & -3 & 0\\ 0 & 1 & 0 & -3\\ -5 & 0 & t^2 & 0\\ 0 & -5 & 0 & t^2 \end{array}\right|\\ &\strut\\ &= \left|\begin{array}{rrr} 1 & 0 & -3\\ 0 & t^2 & 0\\ -5 & 0 & t^2 \end{array}\right| -3\left|\begin{array}{rrr} 0 & 1 & -3\\ -5 & 0 & 0\\ 0 & -5 & t^2 \end{array}\right|\\ &\strut\\ &= t^2\left|\begin{array}{rr} 1 & -3\\ -5 & t^2 \end{array}\right| + 5\left|\begin{array}{rr} 1 & -3\\ -5 & t^2 \end{array}\right|\\ &\strut\\ &= t^2(t^2 -15) + 5(t^2 - 15)\\ &\strut\\ &= (t^2-15)(t^2+5). \end{align*}$$ Of course, this polynomial has $\sqrt{15}$ as a root, but it is not the minimal one that does.

Answer 2

Yes, algebraic numbers are closed under sum and product. Hint: represent algebraic numbers by matrix eigenvalues. Suppose $\rm b$ and $\rm c$ are algebraic integers. Find a nonzero vector $\rm v$ and two integer matrices $\rm B$ and $\rm C$ with $\rm Bv = bv,\ Cv = cv.\:$ Then $\rm \:b+c,\: bc\:$ are eigenvalues of $\rm\: B+C,\: BC\:$ resp.

$$\rm (B+C)v = Bv + Cv = bv + cv = (b+c)v $$ $$\rm \ \ \ \ (B C)v = B(Cv) = B(cv) = cBv = bc v $$

For further details see here.