Prove that if $a +b\sqrt{c}$ is a root of a polynomial in $\mathbb Z$ then $a-b\sqrt{c}$ is also a root of a polynomial in $\mathbb Z$.

Question

Prove that if $a +b\sqrt{c}$ is a root of a polynomial in $\mathbb{Z},$ then $a-b\sqrt{c}$ is also a root of a polynomial in $\mathbb{Z}$. a,b, and c are all integers and c is not a perfect square.

I tried to approach the question similarly to how I'd show the conjugate of a root of a polynomial is also a root of the polynomial, but the calculations aren't coming out cleanly, leading me to believe I'm missing something. Is there some trick or insight I'm missing to make this proof? Thanks.

Answer 1

Hint $ $ Conjugation $\rm\:x\mapsto \bar x\:$ is a ring hom so preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products},\:$ and $\rm\:\color{#0a0}{fixes\ coeff's}\in\Bbb Q.\:$ Therefore, by induction, it preserves polynomial terms $\rm\ \overline{f(w)} = f(\overline w),\ \ f(x)\in\Bbb Q[x],\ $ since such polynomial terms are compositions of said basic operations. $ $ More explicitly

$\!\!\!\begin{eqnarray} \rm \overline{f(w)} &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in R\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in R \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \ \ \forall\ a\in \Bbb Q\\ &=&\rm\ f(\overline w)\\ \rm So\,\ \ 0 = f(w)\, \ \Rightarrow\,\ 0 = \bar 0 = \overline{f(w)}& =&\ \rm f(\overline w)\quad {\bf QED} \end{eqnarray}$

Remark $\ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, hom's commute with polynomials.

Answer 2

Choose some fixed, non-square $c$. Consider the conjugation map: $$f(a+b\sqrt{c})=a-b\sqrt{c}$$ acting over $\mathbb Q[\sqrt{c}]$. You can prove that this satisfies: $$f(x_1+x_2)=f(x_1)+f(x_2)$$ and $$f(x_1x_2)=f(x_1)f(x_2)$$- that is, it is an automorphism of the ring $\mathbb Q[\sqrt{c}]$. Notice that this definition implicitly uses that $a_1+b_1\sqrt{c}=a_2+b_2\sqrt{c}$ implies that $a_1=a_2$ and $b_1=b_2$, which is why it fails for square $c$.

In particular, since polynomials are built from addition and multiplication - that is, they are defined purely on the properties of the ring (of which $f$ is an automorphism) - it implies that all polynomials $p$ in $\mathbb Z$ (a subring of $\mathbb Q[\sqrt{c}]$ where all points are fixed by $f$ - meaning the coefficients of the polynomial don't get messed around) commute with $f$: $$f(p(x))=p(f(x)).$$ Now, if $p(x)=0$, then this becomes: $$f(0)=p(f(x))$$ $$0=p(f(x)).$$ If $x=a+b\sqrt{c}$, this is exactly what you wanted.

Answer 3

Let $\alpha=a+b\sqrt{c}$ and $f(x) \in \mathbb{Z}[x]$ such that $f(\alpha)=0$. Let $\bar{\alpha}=a-b\sqrt{c}$ and $p(x)=(x-\alpha)(x-\bar{\alpha})=x^2-2ax+a^2-b^2c$. Since $p(x)$ is monic, there exist polynomials $q(x)$, $r(x)$ in $\mathbb{Z}[x]$ such that

$f(x)=p(x)q(x)+r(x)$ where $r(x)=Ax+B$.

We have $0=f({\alpha})=r({\alpha})$. Therefore $A(a+b\sqrt{c})+B=0$. Assume $b$ is not 0. Since $c$ is not a perfect square, we deduce $Aa+B=0$ and $Ab=0$ which imply $A=B=0$. Hence $p(x)$ divides $f(x)$ and $f(\bar{\alpha})=0$.

Answer 4

I use some notions of fields extension here. If $a+b\sqrt{c}\notin\mathbb{Q}$ (i.e. $c$ is not a perfect square), then $\mathbb{Q}\left(a+b\sqrt{c}\right)$ is an extension of $\mathbb{Q}$ , and we have$$\left[\mathbb{Q}\left(a+b\sqrt{c}\right):\mathbb{Q}\right]\geq2\,\,\,\,\,\left(*\right).$$ Now consider the polynomial $P=X^{2}-2aX+a^{2}-b^{2}c\in\mathbb{Q}\left[X\right]$ . We have$$P\left(a+b\sqrt{c}\right)=\left(a^{2}+b^{2}c+2ab\sqrt{c}\right)-2a^{2}-2ab\sqrt{c}+a^{2}-b^{2}c=0$$ thus $a+b\sqrt{c}$ is a root of $P$ . Now, since we have $\left(*\right)$ and $\deg\left(P\right)=2$ , we get that $P$ is the minimal polynomial of $a+b\sqrt{c}$ .Now, we also have$$P\left(a-b\sqrt{c}\right)=\left(a^{2}+b^{2}c-2ab\sqrt{c}\right)-2a^{2}+2ab\sqrt{c}+a^{2}-b^{2}c=0$$ so $a-b\sqrt{c}$ is also a root of $P$.

Finally, let be $Q\in\mathbb{Q}\left[X\right]$ such that $Q\left(a+b\sqrt{c}\right)=0$ ; then, $P\mid Q$ by definition of the minimal polynomial, i.e. there an element $R\in\mathbb{Q}\left[X\right]$ such that $PR=Q$ . Evaluating in $X=a-b\sqrt{c}$ we find $$0=P\left(a-b\sqrt{c}\right)R\left(a-b\sqrt{c}\right)=Q\left(a-b\sqrt{c}\right)$$ and then $a-b\sqrt{c}$ is also a root of $Q$ . If $Q\in\mathbb{Z}\left[X\right]$ , then $Q\in\mathbb{Q}\left[X\right]$ too, and we get the same conclusion.