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Let $\alpha$ be a root of $f(x)=x^3+11x+4$. I'd like to compute the field polynomial of $\gamma=(\alpha+\alpha^2)/2$ (that is, the polynomial whose roots consist of the conjugates of $\gamma$).

First, it's not hard to see that $f(x)$ is irreducible over $\mathbb{Q}$ by the rational roots test. So $\mathbb{Q}[\alpha]$ has degree $3$ over $\mathbb{Q}$ and so the field polynomial for $\gamma$ can have degree at most $3$.

First, we have that $\gamma^2=(\alpha^2+2\alpha^3 +\alpha^4)/2=-5\alpha^2-13\alpha -4$ (reducing exponents using the fact that $\alpha$ is a root of $f(x)=0$.

Now, I'm not sure where to go for here, or if there is perhaps a better to solve this than guess and check. How should I proceed?

ponchan
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  • If $\alpha$ is a root of $f(x)$, of degree $3$, then $-\alpha$ is a root of $(-1)^3f(-x)$. It follows that $(-1)^3f(-x)f(x)$ is even. So, if can be written as $g(x^2)$. Then $\alpha^2$ is a root of $g(x)$. Computing the resultant, $R(z)$ of the polynomials $f(x)$ and $g(z-x)$ viewed as a polynomial in $x$. Observe that $\alpha+\alpha^2$ is a root of $R(z)$. Finally, $(\alpha+\alpha^2)/2$ is a root of $R(2z)$. – MoonLightSyzygy Jan 02 '20 at 18:47
  • The above is a general approach. If you want to continue your approach note that you can reduce some more your expression for $\gamma^2$ if you add $10\gamma$ you get a polynomial of degree $1$ in $\alpha$ on the right. Then compute $\gamma^3$ and reduce again until a polynomial of degree $1$ in $\alpha$. Solve for $\alpha$ (in term of $\gamma$) in both equations and eliminate $\alpha$. – MoonLightSyzygy Jan 02 '20 at 18:53

3 Answers3

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Rewriting the symmetric polynomial $\,f(x)f(y)\,$ in terms of $\,xy\,$ and $\,x\!+\!y\ (=-1)\,$ yields

$x\!+\!y = -1,\, f = x^3\!+\!11x\!+\!4\,\Rightarrow\, f(x)f(y) = (xy)^3\!-\!22(xy)^2\!+\!144(xy)\!-\!32 := g(xy)$

so $\,\color{#c00}{-\frac{1}2}xy = x(x\!+\!1)/2\,$ is a root of $\,-g(\color{#c00}{-2}z)/8 = \bbox[5px,border:1px solid #c00]{z^3+11z^3+36z+4}\,\pmod{\!f(x)}$

Remark $\ $ For completeness, below are details of the simple symmetric rewriting

$\qquad \begin{align} f(x)f(y) = (xy)^3 + 4(\!\underbrace{x^3\!+\!y^3}_{\textstyle \color{#c00}{3xy-1}}\!)+11xy(11\!+\!\underbrace{x^2\!+\!y^2}_{\textstyle \color{#0a0}{1-2xy}}\!)+ 44(\underbrace{x\!+\!y}_{\textstyle -1})+16 \end{align}$

$\begin{align}{\rm using}\ \ x^3\!+\!y^3 &= (x\!+\!y)^3\!-\!3(x\!+\!y)xy,\\ &=\ \ \ \color{#c00}{{-}1\ \, -\, \ 3\,(-1)\,xy}\end{align}$ $ \begin{align}{\rm \&}\ \ x^2\!+\!y^2 &= (x\!+\!y)^2\!-\!2xy,\\ &=\ \ \ \ \ \color{#0a0}{1\ -\ 2xy}\end{align} $

Bill Dubuque
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The way you compute this is you find the matrix representing multiplication by $\gamma$. So we compute

\begin{align} \gamma \cdot 1 &= 0 \cdot 1 + \frac12 \cdot\alpha + \frac12 \cdot \alpha^2 \\ \gamma \cdot \alpha &= -2 \cdot 1 - \frac{11}2\cdot \alpha + \frac12\cdot \alpha^2 \\ \gamma \cdot \alpha^2 &= -2 \cdot 1 - \frac{15}2\cdot \alpha-\frac{11}2\cdot \alpha^2 \end{align}

Then the minimal polynomial of the matrix

\begin{pmatrix} 0 & -2 & -2 \\ \frac12 & -\frac{11}2 & -\frac{15}2 \\ \frac12 & \frac12 & -\frac{11}2 \end{pmatrix}

is the minimal polynomial of $\gamma$.

To see this, let $M(t)$ be the matrix corresponding to multiplication of $t$ in the $\mathbb{Q}$-vector space $\mathbb{Q}[\alpha]$. If $f$ is a polynomial then $$f(M(t)) = M(f(t))$$

So if $f(M(\gamma)) = \pmb{0}$ then $M(f(\gamma)) = \pmb 0$ hence $f(\gamma) = 0$.

Trevor Gunn
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  • Isn't it only true that the minimal polynomial of that matrix is a power of the minimal polynomial of $\gamma$ (in case $\gamma$ doesn't generate $\mathbb Q[\alpha]$ - though I think it works out for this specific $\gamma$)? – Milo Brandt Jan 02 '20 at 23:52
  • The characteristic polynomial of the matrix is its minimal polynomial raised to the power $[K:\mathbb{Q}[\gamma]]$ in general. @MiloBrandt – Trevor Gunn Jan 02 '20 at 23:53
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Three things to note here:

  1. $\gamma \in \mathbb{Q}[\alpha] =: E$, so $F := \mathbb{Q}[\gamma] \subseteq E$.
  2. The degree of the sought-after field polynomial is equal to the degree $[F:\mathbb{Q}]$.
  3. Since $\mathbb{Q} \subseteq F \subseteq E$, we know $[F:\mathbb{Q}]$ divides $[E:\mathbb{Q}] = 3$, so we don't have many options for $[F:\mathbb{Q}]$. Now we ask: when is $[F:\mathbb{Q}] = 1$? Is that true in this case?
Eric Canton
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