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I'm having some trouble finding the minimal polynomial of $ \alpha = \zeta_5 + i \sqrt[4]{5}$ over $\mathbb {Q}$, where $\zeta_5$ is the fifth root of 1 ($\zeta_5 := e^{\frac{2\pi}{5}i}$).

I know that $\mathbb {Q}(\zeta_5)\supset$ $\mathbb {Q}(\sqrt[]{5})$ and obviously $(\sqrt[4]{5})^2=\sqrt[]{5}$. Any idea?

RobPratt
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Saber98
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4 Answers4

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The most natural method (not necessarily the most economical) to compute the minimal polynomial over a base field of an algebraic element $\alpha$ is to compute its conjugates and use their symmetric functions. Let me outline it in your example, for $\alpha=\zeta_5 +i\sqrt [4] 5$ over $\mathbf Q$. This is actually a disguised exercise in Galois theory.

First define $k=\mathbf Q (i)$ and determine $K=k(\alpha)$. Introduce the biquadratic (i.e. galois with group $G\cong C_2 \times C_2$) extension $L=k(\zeta_5, i\sqrt [4] 5)$, whose 3 quadratic subextensions are $k(\zeta_5), k(i\sqrt [4] 5), k(\zeta_5 i\sqrt [4] 5)$. It is easy to show that the $G$-conjugates of $\alpha$ over $k$ are the primitive elements of $K/k$, see e.g. https://math.stackexchange.com/a/3325514/300700. This shows that $L=K$ and gives the symmetric functions of $\alpha$ over $k$, hence its minimal polynomial $f_k (X)$ over $k$, of degree 4.

To get the minimal polynomial $f_{\mathbf Q} (X)$ of $\alpha$ over $\mathbf Q$, it suffices to compute the conjugate $g_k (X)$ of $f_k (X)$ under the complex conjugation which generates Gal($k/\mathbf Q$), to obtain finally $f_{\mathbf Q} (X)=f_k (X)g_k (X)$ .

Erratum: See the discussion below

nguyen quang do
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Erratum: To be reclassified as a comment. I don't know how to proceed.

The elementary symmetric functions $s_i$ of the roots $t_j$ of a polynomial are defined in any course in algebra, see e.g. Lang's book. The most well known are the trace and norm. Up to signs, they give the coefficients of the polynomial. More precisely, if $f(X)=(X-t_1)...(X-t_n)$, then $f(X)=X^n-s_1x^{n-1}+...+(-1)^n s_n$, with $s_1=t_1+...+t_n$ (trace), $s_2=\sum t_i t_j,..., s_n=t_1...t_n$ (norm). The word "symmetric" has an obvious origin.

Community
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nguyen quang do
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CORRECTION: All thanks to მამუკა ჯიბლაძე, for disbelieving a “simplification” that I had erroneously made. The very slightly corrected posting follows, giving a radically different final result.

Let me contribute an approach different to that of Nguyen Quang Do. I’ll write $\zeta$ for $\zeta_5$.
We know that the field $K=\Bbb Q(\zeta+i\sqrt[4]5\,)=\Bbb Q(\zeta,i\sqrt[4]5\,)$ is quartic over $k=\Bbb Q(\sqrt5\,)$, in fact biquadratic, the compositum of the quadratic extensions $k(\zeta)$ and $k(i\sqrt[4]5\,)$ of $k$.

First, let’s find the minimal polynomial of $\zeta+i\sqrt[4]5$ over $k$; this is the most work.

Now, \begin{align} \text{Irr}\bigl(\zeta,k[x]\bigr)&=x^2+\frac{1-\sqrt5}2x+1=f(x)&(1)\\ \text{Irr}\bigl(\zeta+i\sqrt[4]5,k(\sqrt[4]5\,)[x]\bigr)&=f(x-i\sqrt[4]5\,) =g(x)\in k(\sqrt[4]5\,)[x]\,. \end{align} The relation $(1)$ is easily found, and I’ll leave the work to you. Next step is to form $\overline g(x)$ by replacing $i\sqrt[4]5$ by $\overline{i\sqrt[4]5}=-i\sqrt[4]5$, and multiplying to get $g\overline g=h(x)$. This is $$ h(x)=\frac{7 - \sqrt5}2 -4x + \frac{7 +3\sqrt5}2x^2 + (1 - \sqrt5)x^3 + x^4\,, $$ equal to the minimal polynomial for $\zeta+i\sqrt[4]5$ over $k$. Notice that its coefficients are $k$-integers, a reassuring observation.

Finally, form $\overline h(x)$ by replacing $\sqrt5$ by $-\sqrt5$, and multiply: \begin{align} \text{Irr}\bigl(\zeta+i\sqrt[4]5, \Bbb Q[x]\bigr)&=h(x)\overline h(x)\\&= 11 - 28x + 48x^2 - 26x^3 + 14x^5 + 3x^6 + 2x^7 + x^8\,. \end{align} This result is the same as that of მამუკა ჯიბლაძე, as appears in one of their comments. Once again, I give thanks for the disbelief.

Lubin
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  • But is the minimal polynomial of $\zeta+i\sqrt[4]5$ the same? I believe it is different. – მამუკა ჯიბლაძე Feb 12 '20 at 18:17
  • And I, @მამუკაჯიბლაძე, believe it is the same. After all, the minimal polynomial for $i\sqrt[4]5$ over $k$ is $x^2+\sqrt5$, while that for $\sqrt[4]5$ is $x^2-\sqrt5$. Since, in the last step, we multiplied $h_{\sqrt5}(x)$ times its conjugate $h_{-\sqrt5}(x)=\overline h(x)$, we should have at this stage also caught $\zeta+i\sqrt[4]5$ as a root. (This is not a proof, just a plausibility argument. It’s too late at night here for me to actually substitute the quantity of interest into my $h\overline h$.) – Lubin Feb 13 '20 at 03:48
  • I now substituted $\zeta+i\sqrt[4]5$ in the polynomial and, if I am not mistaken, it is not zero. It might be that I fail to simplify the radicals but I also evaluated it numerically in Mathematica and it is about $-1748.66-1451.72i$ – მამუკა ჯიბლაძე Feb 13 '20 at 05:44
  • In fact,$$h_1(x):=f(x-i\sqrt[4]5)f(x+i\sqrt[4]5)=\frac{7-\sqrt5}2-4x+\frac{7+3\sqrt5}2x^2+(1-\sqrt5)x^3+x^4,$$and if $\bar h_1(x)$ is obtained from $h_1(x)$ by replacing $\sqrt5$ with $-\sqrt5$ throughout, then for $h_1(x)\bar h_1(x)$ I get$$11-28 x+48 x^2-26 x^3+14 x^5+3 x^6+2 x^7+x^8.$$ This is zero on $\zeta+i\sqrt[4]5$, if I am not making some mistake. – მამუკა ჯიბლაძე Feb 13 '20 at 06:03
  • Troubled by the discrepancy between $\zeta+\sqrt 5$ and $\zeta+i\sqrt 5$, I checked again ... and found an error! Actually $k(\zeta i\sqrt [4] 5)/k$ is not quadratic. If I adapt my proof and replace $k=\mathbf Q(i)$ by $k=\mathbf Q(\sqrt 5)$, then everything rolls and I recover @Lubin's variant $K=\mathbf Q(\zeta+\sqrt [4] 5)$. But now $k(i)$ is not contained in $K$ (we know the quadratic subextensions of $K/k$), so $K(i)/k$ is triquadratic and contains the extension $L=\mathbf Q(\zeta+i\sqrt [4] 5)$ of the OP. In fact $K$ and $L$ are conjugate in $K(i)$. – nguyen quang do Feb 13 '20 at 11:54
  • It’s morning now here, in Minnesota, and time for me to eat crow. The complaints of @მამუკაჯიბლაძე are fully justified. I will redo the computation, correctly this time. But I have some personal business to take care of first. – Lubin Feb 13 '20 at 14:41
  • @Lubin Curiously enough, your polynomial is also minimal for $\zeta^2+i\sqrt[4]5$, while mine is also minimal for $\zeta^2+\sqrt[4]5$ – მამუკა ჯიბლაძე Feb 13 '20 at 16:20
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    That’s not curious at all,@მამუკაჯიბლაძე. I realized, tossing and turning in bed last night, that since both $\zeta+\sqrt[4]5$ and $\zeta+i\sqrt[4]5$ are in the normal field $\Bbb Q(i,\sqrt[4]5,\zeta_5)$, and both are quadratic over $L=\Bbb Q(i,\sqrt[4]5,)$, any automorphism that takes $\sqrt[4]5$ to $i\sqrt[4]5$ will necessarily take $\zeta+\zeta^{-1}$ to $\zeta^2+\zeta^3$. – Lubin Feb 13 '20 at 18:00
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Troubled by the discrepancy between $\zeta+\sqrt 5$ and $\zeta+i\sqrt 5$, I checked again ... and found an error! Actually $k(\zeta i\sqrt [4] 5)/k$ is not quadratic. If I adapt my proof and replace $k=\mathbf Q(i)$ by $k=\mathbf Q(\sqrt 5)$, then everything rolls and I recover @Lubin's variant $K=\mathbf Q(\zeta+\sqrt [4] 5)$. But now $k(i)$ is not contained in $K$ (we know the quadratic subextensions of $K/k$), so $K(i)/k$ is triquadratic and contains the extension $L=\mathbf Q(\zeta+i\sqrt [4] 5)$ of the OP. In fact $K$ and $L$ are conjugate in $K(i)$.

nguyen quang do
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  • PS: I made again an error of manipulation and posted this comment as ananswer !! Please consider this "new answer" as being deleted. – nguyen quang do Feb 13 '20 at 12:00