Find minimal polynomial of $ \alpha = \zeta_5 + i \sqrt[4]{5}$ over $\mathbb {Q}$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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 $K(i)$.