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I tried to use the Briot-Ruffini method but it didn't work.

The question I need help is: "Prove that, if a polynomial equation with integer coefficients has the irrational number $a+\sqrt{b}$ as a root, with $a,b \in \mathbb{Z} $, $b$ a prime number, so is $a-\sqrt{b}$."

Vinícius
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  • You also may be able to prove using fact that any polynomial with real coefficients can be factors to degree 1 or 2 polynomials. From this since integer coefficient and a+sqrt(b) is a root there must be a degree polynomial in factor it is root of than argue for that factor that it is a-sqrt(b) is also factor – Kamster Jun 29 '14 at 23:31
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    @user159813 "fact that any polynomial with real coefficients can be factors to degree 1 or 2 polynomials" - if by that you mean that any polynomial can be factorised over $\mathbb Q$ into linear and quadratic factors, then that is incorrect (e.g. $X^n - 2$ for all $n$) – Mathmo123 Jun 29 '14 at 23:33

5 Answers5

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If $p(x)$ is a polynomial with integer coefficients, and $a$ is an integer, then $q(x)=p(a+x)$ is also a polynomial with integer coefficients. If $a+\sqrt b$ is a root of $p(x)$, then $q(\sqrt b)=p(a+\sqrt b)=0$. Now any polynomial with integer coefficients can be written in the form $r(x^2)+xs(x^2)$, where $r$ and $s$ also have integer coefficients. If we write $q$ this way, then we have

$$0=q(\sqrt b)=r(b)+s(b)\sqrt b$$

which implies $r(b)=s(b)=0$ if $b$ is not a perfect square. But this implies

$$p(a-\sqrt b)=q(-\sqrt b)=r(b)-s(b)\sqrt b=0$$

hence $a-\sqrt b$ is also a root of $p(x)$.

Barry Cipra
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Hint: $a+\sqrt b$ is a root of $(x-(a+\sqrt b))(x - (a-\sqrt b)) = x^2 -2ax + (a^2-b)$.

(This is the minimal polynomial of $a + \sqrt b$ over $\mathbb Z$, and hence, it divides any other polynomial that has $a + \sqrt b$ as a root)

Mathmo123
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Another way, similar to Barry Cipra's method:

Assume that $P$ is a polynomial with rational coefficients and let $a$ and $b$ be rationals such that $\sqrt{b}$ is irrational such that $P(a+\sqrt{b})=0$.

By the binomial theorem, $P(a+\sqrt{b})+P(a-\sqrt{b})$ is rational. To see why, write $P(x)=\sum_{k=0}^nc_kx^k$, and expand $P(a \pm \sqrt{b})$; when you sum $P(a+\sqrt{b})$ and $P(a-\sqrt{b})$, the terms containing $\sqrt{b}$ cancel.

By similar reasoning $P(a+\sqrt{b})-P(a-\sqrt{b})$ is some rational multiple of $\sqrt{b}$.

But $P(a+\sqrt{b})=0$, thus $P(a-\sqrt{b})$ is both rational and some rational multiple of $\sqrt{b}$. Hence $P(a-\sqrt{b})=0$.

math_lover
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Hint $ $ The set $I$ of $\,f\in\Bbb Q[x]\,$ with root $\,w = a+\sqrt b\,$ is an ideal, i.e. $I$ is closed under subtraction & multiplication by any $\,g\in \Bbb Q[x].\,$ Thus by the Euclidean algorithm $\,I = (g)\,$ is principal, generated by any min degree $\,g\in I\,$ (else $\,0\ne f\ {\rm mod}\ g = f - q\:\! g \in I\,$ & has degree $< \deg g)$.

By $\,\sqrt b\not\in\Bbb Q,\,$ a min degree $\,g \in I\,$ has degree $\,2,\,$ e.g. $\,g = (x-w)(x-\bar w),$ $\ \bar w = a-\sqrt b = $ conjugate of $\,w.\,$ Hence $\ f(w) = 0\iff f\in (g)\iff g\mid f\iff w,\bar w\,$ are roots of $\,f.$

Remark $ $ The same idea works generally to show that ideals in Euclidean domains (i.e. domains enjoying division with "smaller" remainder) are principal, generated by any element of minimal Euclidean size (here size = degree, and the generator $\,g\,$ is called a "minimal polynomial" for $\,w)$.

Bill Dubuque
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  • Why do you know that already degree 2 will be enough for having $a+\sqrt b$ as a root? – 355durch113 May 18 '16 at 06:03
  • @355durch113 $\ w,w',$ are roots of $\ g = (x-w)(x-w') = x^2-(w+w')x +ww'\in\Bbb Q[x],$ since its coef's are rational, by $,w+w' = 2a,,$ and $,ww' = a^2-b,,$ i.e. $w$ has rational trace and norm. – Bill Dubuque May 18 '16 at 14:46
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Conjugation $\mathbb{Q}(\sqrt{b}) \to \mathbb{Q}(\sqrt{b}):z= x+y\sqrt{b} \to \bar{z}=x-y\sqrt{b}$ is a field automorphism that fixes $\mathbb{Q}.$ So if $z\in \mathbb{Q}(\sqrt{b})$ is the root of a polynomial with rational coefficients i.e. $\exists \ a_i\in \mathbb{Q}$ such that $$a_n z^n + a_{n-1}z^{n-1} + \ldots + a_1 z +a_0=0,$$

then conjugating both sides of the equation shows that $\bar{z}$ is a root of the same polynomial.

A similar result is the Conjugate Root Theorem. More generally, if $K\subseteq L$ is a field extension and $\varphi:L\to L$ is an automorphism of $L$ that fixes $K,$ then $$ l \in L \text{ is a root of } p\in K[X] \implies \varphi(l) \text{ is a root of } p\in K[X]. $$

This is used frequently when you study Field/Galois theory.

Ragib Zaman
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