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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 $a-\sqrt{b}$."

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  • $\begingroup$ 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 $\endgroup$
    – Kamster
    Jun 29, 2014 at 23:31
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    $\begingroup$ @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 $n$) $\endgroup$
    – Mathmo123
    Jun 29, 2014 at 23:33

5 Answers 5

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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)

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    $\begingroup$ Nice answer, you should say irreducible instead of minimal $\endgroup$ Jun 29, 2014 at 23:14
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    $\begingroup$ One should explain how it being "minimal" plays any role here. $\endgroup$ Jun 29, 2014 at 23:14
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    $\begingroup$ I guess that wasn't clear. (Although being the min poly is equivalent to being irreducible) $\endgroup$
    – Mathmo123
    Jun 29, 2014 at 23:16
  • $\begingroup$ @Mathmo123, technically the minimal polynomial is the monic irreducible, but yes. $\endgroup$
    – Kaj Hansen
    Jun 29, 2014 at 23:17
  • $\begingroup$ Right again! It's not been my day... $\endgroup$
    – Mathmo123
    Jun 29, 2014 at 23:19
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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 $p(x)$.

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  • $\begingroup$ Thanks! That is what I need. $\endgroup$
    – Vinícius
    Jun 29, 2014 at 23:40
  • $\begingroup$ Good solution ! $\endgroup$
    – Silva
    Apr 22, 2016 at 21:52
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    $\begingroup$ How can I know this "any polynomial with integer coefficients can be written in the form $r(x^2)+xs(x^2)$" for sure? $\endgroup$
    – space
    Jan 26, 2018 at 16:40
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    $\begingroup$ @Helena, consider an example: $$x^6+2x^5+3x^4+4x^3+5x^2+6x+7=(x^6+3x^4+5x^2+7)+x(2x^4+4x^2+6)$$ in which case $r(x)=x^3+3x^2+5x+7$ and $s(x)=2x^2+4x+6$. Does that clarify things? $\endgroup$ Jan 26, 2018 at 19:07
  • $\begingroup$ @ Barry Cipra — yes, now I see how it can be like that, but does that come from a formula or a theorem etc.? $\endgroup$
    – space
    Jan 29, 2018 at 0:25
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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.

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  • $\begingroup$ Nice answer!... $\endgroup$ Dec 27, 2019 at 9:09
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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 $P(a-\sqrt{b})=0$.

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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. is closed under subtraction & multiplication by any $\,g\in \Bbb Q[x].\,$ So by the Euclidean algorithm $\,I = (g)$ is principal, generated by any element of minimal degree (else $\,0\ne f\ {\rm mod}\ g = f - q\:\! g \in I\,$ and has degree smaller than $\,g).$

By $\,\sqrt b\not\in\Bbb Q,\,$ a min degree $\,g \in I\,$ has degree $\,2,\,$ e.g. $\,g = (x-w)(x-w'),$ $\, w' = a-\sqrt b = $ conjugate of $\,w.\,$ Hence $\ f(w) = 0\iff f\in (g)\iff g\mid f\iff w,w'\,$ 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 (above the size is the polynomial degree)

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  • $\begingroup$ Why do you know that already degree 2 will be enough for having $a+\sqrt b$ as a root? $\endgroup$ May 18, 2016 at 6:03
  • $\begingroup$ @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. $\endgroup$ May 18, 2016 at 14:46

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