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.

Solution 1:

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.

Solution 2:

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 $C_p^2$.