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Let $a,b \in \mathbb{Q}$ and $d \neq 0,1$ be a square free integer.

Define $\overline{a + b\sqrt{d}} = a - b\sqrt{d}$

If $f \in \mathbb{Z}[x]$ show that: $f(\overline{\alpha}) = \overline{f(\alpha)}$

The notation may be somewhat confusing. Prior to this we had defined a set: \begin{equation*} A = \{a + b\sqrt{d} : a,b \in \mathbb{Q}\} \cap \overline{\mathbb{Z}} \end{equation*} where $\overline{\mathbb{Z}}$ is the algebraic closure of $\mathbb{Z}$.

Here is my working so far:

A polynomial $f$ in $\mathbb{Z}[x]$ has the form:

$$ f(x) = \sum_{i=0}^n a_ix^i $$

I think we might require that $f$ be monic, so let $|a_n| = 1$.

Now, by the binomial theorem:

$$(a - b\sqrt{d})^i = \sum_{k=0}^i \binom{i}{k} a^{i-k}(-b\sqrt{d})^k$$

and as such we may write:

$$ f(\overline{\alpha}) = \sum_{i=0}^n a_i\sum_{k=0}^i \binom{i}{k} a^{i-k}(-b\sqrt{d})^k $$

But this becomes real messy real fast! I don't even know how to begin calculating $\overline{f(\alpha)}$. Could someone provide some advice for showing $f(\overline{\alpha}) = \overline{f(\alpha)}$?

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You need something more structural. First show that for $z$ and $w$ of your form, $\overline{z+w}=\overline{z}+\overline{w}$, and the same thing for multiplication. It all flows from that. –  Lubin Oct 14 '12 at 2:58
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3 Answers

up vote 2 down vote accepted

As said in comments, what you need to prove first is that $x\mapsto \overline x$ is a ring homomorphism (that is, $\overline 1=1$, $\overline x+\overline y=\overline{x+y}$, and $\overline x\cdot \overline y=\overline{x\cdot y}$), and also that conjugation is the identity on $\mathbb Z$.

Once you have this, you can prove by induction that $\overline{\alpha}^n=\overline{\alpha^n}$ and then $a_n\overline\alpha^n = \overline{a_n\alpha^n}$ because the coefficient is an integer. Finally, the sum of the conjugated terms is the conjugate of the sum of the terms.

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Induct on degree of $\rm\ \ f(x)\, =\, a + x\,g(x)\ \ $ as follows

$$\begin{eqnarray}\rm \overline{f(x)} &=&\rm \overline{a + x\,g(x)} \\ &\,=\,&\rm a + \overline{x\, g(x)} \ \ \ by\ \ \ \overline{y+z} = \overline{y} + \overline{z},\ \ \overline n = n,\ \ for\ n\in\Bbb Z \\ &=&\rm a + \overline{x}\, \overline{g(x)} \ \ \ by\ \ \ \overline{y\, *\, z} = \overline{y} \,* \overline{z}, \\ &=&\rm a + \overline{x}\, g(\overline x)\ \ \ by\ induction,\ \ since\ \ deg\ g < deg\ f\\ &=&\rm f(\overline x) \end{eqnarray}$$

Remark $\ $ The proof uses only that $\rm\:z\to \bar z\:$ is a ring homomorphism that fixes the coefficients.

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Given that $f(x)=\sum_{i=0}^da_ix^i$ just compute $$ \overline{f(\alpha)}=\overline{\sum_{i=0}^da_i\alpha^i}= \sum_{i=0}^d\overline{a_i\alpha^i}=\sum_{i=0}^d\bar{a}_i\bar{\alpha}^i= \sum_{i=0}^da_i\bar{\alpha}^i $$ since $a_i\in\Bbb Z$. Finally, the last expression is just $f(\bar\alpha)$.

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