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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

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Explicitly produce an equation in rational coefficients for which this is a root. –  user21436 Mar 10 '12 at 22:13
    
@KannappanSampath - is this involve guessing? i hate guessing –  Victor Mar 10 '12 at 22:14
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Any number obtained by a finite combination of algebraic operations (addition, multiplication, root extraction) is algebraic. –  Arturo Magidin Mar 10 '12 at 22:15
    
@Victor: No, it doesn't involve guessing; in this case, it is pretty easy: if $a=\sqrt{5}+\sqrt{7}$, then squaring gives $a^2 = 12+2\sqrt{35}$; so $a^2-12 = 2\sqrt{35}$. So $(a^2-12)^2 = 140$. Hence, $a$ is a root of $(x^2-12)^2 - 140$. –  Arturo Magidin Mar 10 '12 at 22:17
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@Victor I think you need to rewrite this post. Define what is given, what is unknown, what is the process to find the unknown, and what exactly do you need to prove is algebraic. –  user2468 Mar 10 '12 at 22:50

2 Answers 2

up vote 6 down vote accepted
  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.

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Nice answer. +1. : ) –  user21436 Mar 10 '12 at 23:08
    
@Victor: Sigh: I did the example with $\sqrt{3}+\sqrt{5}$ instead of $\sqrt{5}+\sqrt{7}$ (that's what I get for not double checking). I hope the method is clear anyway. –  Arturo Magidin Mar 10 '12 at 23:37

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.

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