If $a+\sqrt{b}$ is a root of a polynomial equation with integer coefficients, so is $a-\sqrt{b}$

Question

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 $a-\sqrt{b}$."

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
@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

Mathmo123 · Answer 1 · 2014-06-29T23:28:54.003

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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)

edited Jun 29 '14 at 23:28

answered Jun 29 '14 at 23:12

Mathmo123

23,018

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Nice answer, you should say irreducible instead of minimal – Rene Schipperus Jun 29 '14 at 23:14
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One should explain how it being "minimal" plays any role here. – Bill Dubuque Jun 29 '14 at 23:14
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I guess that wasn't clear. (Although being the min poly is equivalent to being irreducible) – Mathmo123 Jun 29 '14 at 23:16
@Mathmo123, technically the minimal polynomial is the monic irreducible, but yes. – Kaj Hansen Jun 29 '14 at 23:17
Right again! It's not been my day... – Mathmo123 Jun 29 '14 at 23:19
No worries @Mathmo123. Great answer regardless. +1 – Kaj Hansen Jun 29 '14 at 23:23
@Mathmo123: a polynomial can't be “minimal”, only “irreducible” or “minimal polynomial of something”. Irreducibility is a property of a polynomial, while minimality is a property of a polynomial and an element. – tomasz Jun 30 '14 at 04:15
OK fine... but now that becomes a matter of bad grammar on my part. It should be fine now – Mathmo123 Jun 30 '14 at 09:44

score 7 · Answer 2 · answered Jun 29 '14 at 23:38

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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 $p(x)$.

answered Jun 29 '14 at 23:38

Barry Cipra

79,832

Thanks! That is what I need. – Vinícius Jun 29 '14 at 23:40
Good solution ! – Silva Apr 22 '16 at 21:52
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How can I know this "any polynomial with integer coefficients can be written in the form $r(x^2)+xs(x^2)$" for sure? – space Jan 26 '18 at 16:40
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@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? – Barry Cipra Jan 26 '18 at 19:07
@ Barry Cipra — yes, now I see how it can be like that, but does that come from a formula or a theorem etc.? – space Jan 29 '18 at 00:25

math_lover · Answer 3 · 2020-04-13T21:36:02.677

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 $P(a-\sqrt{b})=0$.

Bill Dubuque · Answer 4 · 2020-10-30T23:53:59.647

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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)

edited Oct 30 '20 at 23:53

answered Jun 29 '14 at 23:24

Bill Dubuque

272,048

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

Ragib Zaman · Answer 5 · 2014-06-30T02:31:51.170

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.

If $a+\sqrt{b}$ is a root of a polynomial equation with integer coefficients, so is $a-\sqrt{b}$

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