13
$\begingroup$

Prove that if the number $x$ is algebraic, then $x^2$ is also algebraic. I understand that an algebraic number can be written as a polynomial that is equal to $0$. However, I'm baffled when showing how $x^2$ is also algebraic.

$\endgroup$
  • 1
    $\begingroup$ If a polynomial exists as $p(x)=ax^n+bx^{n-1}+...+q = 0$ then what can you say about plugging in $x=y^2$? $\endgroup$ – abiessu Nov 17 '14 at 23:09
  • 6
    $\begingroup$ @abiessu: I think that shows that if $x^2$ is algebraic then $x$ is... $\endgroup$ – Ben Millwood Nov 17 '14 at 23:10
  • $\begingroup$ @abiessu: Would p(y^2) = ay^(2n) + by^(2n-2) +...+ q = 0? $\endgroup$ – Bob Nov 17 '14 at 23:17
  • 1
    $\begingroup$ @abiessu, I don't understand how that will help? Can you please elaborate? $\endgroup$ – Bob Nov 17 '14 at 23:32
  • 1
    $\begingroup$ I was wrong, it was a foolish suggestion. There are some good answers below. $\endgroup$ – abiessu Nov 18 '14 at 5:33
26
$\begingroup$

We have $P(x)=0$, where $P$ is some rational polynomial (that is, the coefficients are rational numbers). Break $P$ up into the terms with odd exponents and the terms with even exponents. For example, if $P(x)=x^4+x^3+5x^2+x+4$, then we would break it up as $(4+5x^2+x^4)+(x+x^3)$. The term with even exponents can be viewed as a polynomial in $x^2$: $4+5x^2+x^4=4+5x^2+(x^2)^2$ . The term with odd exponents can be viewed $x$ times a polynomial in $x^2$: $x+x^3=x(1+x^2)$. We now have an identity of the form:

$$P(x)=A(x^2)+xB(x^2)=0$$

Where $A$ and $B$ have rational coefficients. We almost have a rational polynomial which is $0$ at $x^2$, we just have to get rid of that $x$ infront of $B(x^2)$. We can do that like so:

$$(A(x^2)+xB(x^2))(A(x^2)-xB(x^2))=A(x^2)^2-x^2B(x^2)^2=0$$

And so the polynomial $A(X)^2 - XB(X)^2$ admits $x^2$ as a root, and of course has rational coefficients.

$\endgroup$
  • $\begingroup$ So, $A(x^2)=\frac12(P(x)+P(-x))$ and $xB(x^2)=\frac12(P(x)-P(-x))$? $\endgroup$ – Akiva Weinberger Nov 18 '14 at 3:33
  • 1
    $\begingroup$ Now that is one cute answer, +1! $\endgroup$ – Robert Lewis Nov 18 '14 at 4:51
8
$\begingroup$

Here's another possibility: if $p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0$ is a polynomial, we can form the companion matrix $$ C_p=\left[\begin{array}{ccccc}0&0&\ldots&0&-a_0\\ 1&0&\ldots&0&-a_1\\ 0&1&\ldots&0&-a_2\\ \vdots&&\ddots&0&\vdots\\ 0&0&\ldots&1&-a_{n-1} \end{array}\right]. $$ It is constructed so that $p(x)$ is the characteristic polynomial of $C_p$.

Suppose $x$ is a root of $p(x)$. Then $x$ is an eigenvalue of $C_p$, implying $x^2$ is an eigenvalue of of $C_p^2$. This means $x^2$ is a root of the characteristic polynomial of $C_p^2$.

$\endgroup$
6
$\begingroup$

Note that $x$ is algebraic over a field $F$ if and only if the field $F(x)$ is a finite dimensional vector space over $F$ (see explanation below). Since $F(x^2)$ is contained in $F(x)$, the claim follows.

We now explain in more details. $F(x)/F$ is a field extension, and so $F(x)$ is automatically a vector space over $F$. If for some $p\in F[t]$ we have $p(x)=0$, it means that $x^d\in\mathrm{span}(1,x,x^2,\ldots,x^{d-1})$, where $d$ is the degree of $p$. Multiplying by $x$ we see that $x^{d+1}\in\mathrm{span}(x,\ldots,x^d)\subset\mathrm{span}(1,\ldots,x^{d-1})$, and by induction, all the powers of $x$ are in $\mathrm{span}(1,\ldots,x^{d-1})$, hence $F(x)$ is finite dimensional over $F$.

Conversly, if $F(x)$ is finite dimensional over $F$, let $d$ denote its dimension, and since $1,x,\ldots,x^d$ are $d+1$ elements, they are linearly dependent over $F$, hence $x$ is a zero of some polynomial with coefficients in $F$.

$\endgroup$
  • $\begingroup$ Is it possible if you could prove this question through polynomials? $\endgroup$ – Bob Nov 17 '14 at 23:26
  • $\begingroup$ @Bob I added a proof of my main argument. Perhaps it answers your question. $\endgroup$ – Amitai Yuval Nov 18 '14 at 10:08
2
$\begingroup$

Slightly changing to a more comfortable notation, let $\alpha$ be an algebraic number that is a root of the polynomial equation $f(x)=0$ having integer coefficients. Now separate the odd powers and even powers in $f(x)$ calling them $f_{\rm odd}(x)$ and $f_{\rm even}(x)$. So $\alpha $ satisfies $f_{\rm odd}(\alpha)=-f_{\rm even}(\alpha)$. Now square both sides and note that both sides will involve only even powers of $\alpha $, and so $\alpha^2$ is algebraic being the root of the equation $f_{\rm odd}(x)^2 -f_{\rm even}(x)^2=0$ where we can substitute $\alpha^2$ in place of $x^2$.

$\endgroup$
  • $\begingroup$ Duplicate of Jack's prior answer $\endgroup$ – Bill Dubuque Oct 20 '18 at 23:47
2
$\begingroup$

Basically $P(x)\cdot P(-x)= Q(x^2)$, a polynomial in $x^2$. A similar trick using $n$-th roots of $1$ allows us to get $P(x) \cdot P_1(x) = Q(x^n)$ for some $Q$ so an equation for $x^n$. To get an equation satisfied by a general $\phi(x)$ ( $\phi(x) = x^2$ in the above example) we can use the companion matrix as @Julian Rosen: wrote in his answer. Here is a concrete example: We know that $2 x^5 + x-2=0$. We want the equation satisfied by $\phi(x) = x^3 - x^2 -1$.

Consider the companion matrix (see http://en.wikipedia.org/wiki/Companion_matrix) \begin{eqnarray} A = \left( \begin{array}{ccccc} 0 & 0 & 0 &0 &1 \\ 1 & 0 & 0 &0 &-\frac{1}{2}\\ 0 & 1 & 0 &0 &0\\ 0 & 0 & 1 &0 &0\\ 0 & 0 & 0 &1 &0\\ \end{array} \right) \end{eqnarray}

Take $\phi(A) = A^3 - A^2 -I_5 = B$ where \begin{eqnarray} B = \left( \begin{array}{ccccc} -1 & 0 & 1 &-1 &0 \\ 0 & -1 & -\frac{1}{2} &\frac{3}{2} &-1\\ -1 & 0 &-1 &-\frac{1}{2} &\frac{3}{2}\\ 1 & -1 & 0 &-1 &-\frac{1}{2}\\ 0 & 1 & -1 &0 &-1\\ \end{array} \right) \end{eqnarray}

The characteristic polynomial of $B$ is $$R(t) =t^5+5 t^4+16 t^3+\frac{61}{2} t^2+\frac{271}{8}t+\frac{127}{8}$$ and one checks that $$R[x^3 - x^2 -1] =\frac{1}{8} (2 x^5+x-2) (4 x^10-20 x^9+40 x^8-40 x^7+18 x^6+10 x^5-16 x^4-12 x^3+23 x^2-x-2)$$ that is, $x^3 - x^2 -1$ is a root of $R(\cdot)$.

Or, with numerics, the equation $2 x^5 + x-2=0$ has a unique real solution $0.8890618537791..$. We get $x^3 - x^2 -1 = -1.0876889476...$ and $R(-1.0876889476...) \simeq 0$

$\bf{Added}$ This method can be used to find equations for polynomial expressions involving several algebraic numbers $x_1$, $x_2$, $\ldots$ each satisfying a given equation. In that case we work with Kronecker products. This is finiteness effective; also see @Amitai Yuval: answer.

$\endgroup$
1
$\begingroup$

Let $f(t)$ be the minimal polynomial of $x$. We can show that $y = g(x)$ is algebraic for any polynomial $g$, using the resultant:

$$ h(s) = \operatorname{Res}_t(f(t), s - g(t)) $$

The resultant is a polynomial in $y$ whose coefficients are taken from the same ring as the ceofficients of $f$ and $g$.

Among the various formulas for the resultant are (for some constant $c$ that's irrelevant)

$$ h(s) = c \prod_{\alpha} (s - g(\alpha)) $$

where $\alpha$ ranges over all of the roots of $f(t)$. Since $x$ is one of the roots of $f(t)$, it's clear from this formula that $y$ is a root of $h(s)$.

In short, the product of all of the conjugates of $s - y$ is a polynomial in $s$ with rational coefficients.

There is another formula (for another irrelevant constant $d$)

$$ h(s) = d \prod_{\beta(s)} f(\beta(s)) $$

where $\beta(s)$ ranges over all of the roots for $t$ of the equation $s - g(t)$. For the case of $g(t) = t^2$, we recover the formula that appears in the other answers:

$$ h(s) = f(\sqrt{s}) f(-\sqrt{s}) $$

which is basically the formula appearing in this answer.

$\endgroup$

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.