If the number $x$ is algebraic, then $x^2$ is also algebraic

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.

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

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.

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

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 $C_p^2$.

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 $F$.

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

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 $x^2$.

• Duplicate of Jack's prior answer – Bill Dubuque Oct 20 '18 at 23:47

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.

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.