The unit norm index and $p$-class group in certain degree $\ell$ extensions

Since $K_{n}$ is a totally imaginary field of degree $2^{n+1}$ over $\mathbb{Q}$, it has no real embeddings ($r_{1}=0$) and $2^{n}$ pairs of complex embeddings ($r_{2}=2^{n}$). By Dirichlet’s unit theorem, $K_{n}$ has

fundamental units.

Let $E_{n}$ be the unit group of $K_{n}$ and let $U_{n}$ be the units of $K_{n}$ modulo $\ell^{th}$ powers.

We also have $\textrm{Gal}(K_{n}/\mathbb{Q})\simeq D_{n}$. Let $\sigma$ and $\tau$ be generators of the Galois group where $\sigma$ has order $2^{n}$ and $\tau$ has order 2. Let $\tau$ be complex conjugation under some embedding $K_{n}\hookrightarrow\mathbb{C}$. We then have the following relation:

Recall that we use $U_{n}$ to represent the units of $K_{n}$ modulo $\ell^{th}$ powers. We now need to understand the structure of the relative units of $K_{n}$ modulo $\ell^{th}$ powers, which we define as

We also need the following result on the structure of the relative unit group.

Finally, we have the following proposition showing that the only ‘new’ units introduced at the $n^{th}$ layer are the relative units.

We need to understand the behavior of the primes in the anti-cyclotomic extension.

We also require the following lemma relating the $\sigma$ action on the primes of $K_{n}$ to the $\tau$ action.

From this point forward, we let our imaginary quadratic field $K_{0}$ be $\mathbb{Q}(i)$ in order to simplify the already technical proofs. This choice has several important consequences. We will be interested in primes that are inert in $K_{0}$ (so that they split completely in $K_{\infty}/K_{0}$). For other base fields, we would have different congruence conditions determining which primes ramify in $L/\mathbb{Q}$. We would also have different congruence conditions determining which primes are fixed by elements of the Galois group (see Proposition 2.7). Finally, we will also see that the choice of $K_{0}=\mathbb{Q}(i)$ means that $K_{1}$ is a cyclotomic field; this is not true in general.

The matrices of norm residue symbols will turn out to have a special form: they will be skew circulant matrices.

The associated polynomial to a skew circulant matrix is

Then for $0\leq i\leq n-1$, each row of the matrix is given by

We’re now ready to prove the theorem.

There are $2^{n-1}$ generators of the relative units modulo $\ell^{th}$ powers in $K_{n}$. By Corollary 2.2, at least one relative unit is fixed by $\tau$. By Proposition 2.3, we may choose $u$ to be a generator that is fixed by $\tau$ and then

is a set of generators for $U_{n}^{rel}$. These are the units in $K_{n}$ that have norm 1 in $K_{n-1}$ modulo $\ell^{th}$ powers. By Proposition 2.4 and Lemma 4.1, these are the only units we need to consider, as we can assume we already know if units from fields lower in the tower are norms.

A unit $u\in K_{n}$ is the norm of an element in $L_{n}$ if and only if its norm residue symbols for each prime that ramifies in $L_{n}/K_{n}$ are trivial. Let $u_{j}=\sigma^{j}u$. Observe that this implies

since $\sigma^{2^{n-1}}u=u^{-1}$ (Proposition 2.3).

We can consider each set of conjugate primes separately, so we’ll first look only at primes that all lie over the same rational prime $p$. Write these primes as

We know there are $2^{n}$ primes in $K_{n}$ lying over $p$, using Lemma 2.5 and our assumption that $p\equiv 3\text{ mod }4$.

We will make use of the following lemma:

We’ll address four cases independently: we treat $\ell\equiv 3\text{ mod }8$ and $\ell\equiv 5\text{ mod }8$ separately, and we treat $p\equiv 7\text{ mod }8$ and $p\equiv 3\text{ mod }8$ separately.

$\bullet\hskip 10.00002ptp\equiv 7\text{ mod }8$, $\ell\equiv 3\text{ mod }8$:

If $p\equiv 7\text{ mod }8$, then by Proposition 2.7 at least one of the primes above $p$ is fixed by $\tau$. Let $\mathfrak{p}$ be the fixed prime: $\tau\mathfrak{p}=\mathfrak{p}$. Then

$$\tau\sigma^{j}\mathfrak{p}=\sigma^{-j}\tau\mathfrak{p}=\sigma^{-j}\mathfrak{p}.$$

Construct a matrix $(a_{ij})$ using the norm residue symbols:

$\displaystyle\left(\frac{u_{j}}{\mathfrak{p}_{i}}\right)\equiv u_{j}^{(N\mathfrak{p}_{i}-1)/\ell}\equiv\omega^{a_{ij}}\text{ mod }\mathfrak{p}_{i}$

where $\omega$ is a fixed primitive $\ell^{th}$ root of unity. We extend $\tau$ and $\sigma$ such that $\omega$ is fixed by both. Since each $\mathfrak{p}_{i}$ is inert in $K_{0}/\mathbb{Q}$ and totally split in $K_{n}/K_{0}$, $N\mathfrak{p}_{i}=p^{2}$. Since we’re considering the powers of $\ell^{th}$ roots of unity, the matrix entries $a_{ij}$ are in $\mathbb{F}_{\ell}$.

If the matrix is trivial, then all of the norm residue symbols are trivial, and therefore all of the units are norms modulo $\ell^{th}$ powers. If the matrix has full rank, then none of the units are norms modulo $\ell^{th}$ powers.

First consider $\mathfrak{p}_{0}=\mathfrak{p}$. Write

$$\left(\frac{u_{j}}{\mathfrak{p}}\right)=\omega^{a_{j}}.$$

For $0\leq j\leq 2^{n-1}-1$, apply $\tau$, using $\tau u_{j}=u_{2^{n-1}-j}^{-1}$, $\tau\omega=\omega$ and $\tau\mathfrak{p}=\mathfrak{p}$:

$\displaystyle\omega^{a_{j}}=\left(\frac{u_{j}}{\mathfrak{p}}\right)=\tau\left(\frac{u_{j}}{\mathfrak{p}}\right)=\left(\frac{u_{2^{n-1}-j}^{-1}}{\mathfrak{p}}\right)=\omega^{-a_{2^{n-1}-j}}\implies a_{j}=-a_{2^{n-1}-j}.$

We want to pay special attention to what happens when $j=2^{n-1}-j$, or $j=2^{n-2}$. Then

$$a_{2^{n-2}}=-a_{2^{n-2}}\implies a_{2^{n-2}}=0.$$

We can use the above relations to write the first row of the matrix $(a_{ij})$ in terms of $a_{j}$, $0\leq j<2^{n-2}$:

$$\begin{pmatrix}a_{0}&a_{1}&\ldots&a_{2^{n-2}-1}&0&-a_{2^{n-2}-1}&\ldots&-a_{1}\end{pmatrix}.$$

To get the next row, we can apply $\sigma$. Then

$$\omega^{a_{j}}=\left(\frac{u_{j}}{\mathfrak{p}}\right)=\sigma\left(\frac{u_{j}}{\mathfrak{p}}\right)=\left(\frac{u_{j+1}}{\sigma\mathfrak{p}}\right)$$

for $0\leq j<2^{n-1}-1$ and

$$\omega^{a_{1}}=\left(\frac{u^{-1}_{2^{n-1}-1}}{\mathfrak{p}}\right)=\sigma\left(\frac{u^{-1}_{2^{n-1}-1}}{\mathfrak{p}}\right)=\left(\frac{u_{0}}{\sigma\mathfrak{p}}\right).$$

If we continue in this manner, we can construct a $2^{n}\times 2^{n-1}$ matrix, where the top half and bottom half are both skew circulant:

(4.1)

$$\begin{pmatrix}a_{0}&a_{1}&\ldots&a_{2^{n-2}-1}&0&-a_{2^{n-2}-1}&\ldots&-a_{1}\\ a_{1}&a_{0}&\ldots&a_{2^{n-2}-2}&a_{2^{n-2}-1}&0&\ldots&-a_{2}\\ a_{2}&a_{1}&\ldots&a_{2^{n-2}-3}&a_{2^{n-2}-2}&a_{2^{n-2}-1}&\ldots&-a_{3}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ -a_{1}&-a_{2}&\ldots&0&a_{2^{n-2}-1}&a_{2^{n-2}-2}&\ldots&a_{0}\\ -a_{0}&-a_{1}&\ldots&-a_{2^{n-2}-1}&0&a_{2^{n-2}-1}&\ldots&a_{1}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ a_{2}&a_{3}&\ldots&-a_{2^{n-2}-1}&-a_{2^{n-2}-2}&-a_{2^{n-2}-3}&\ldots&-a_{1}\\ a_{1}&a_{2}&\ldots&0&-a_{2^{n-2}-1}&-a_{2^{n-2}-2}&\ldots&-a_{0}\\ \end{pmatrix}$$

If $n=2$, we have the matrix

$$\begin{pmatrix}a_{0}&0\\ 0&a_{0}\\ -a_{0}&0\\ 0&-a_{0}\end{pmatrix}$$

which has rank 0 if $a_{0}=0$ and rank $2$ if $a_{0}\neq 0$. This means that either both units are local norms at all four primes or neither of them are.

Now we consider $n>2$. By Proposition 4.4, the top half of the matrix has eigenvectors

$$(1,\zeta,\ldots,\zeta^{2^{n-1}-1})$$

where $\zeta$ is one of the $2^{n-1}$ distinct solutions to $x^{2^{n-1}}+1=0$. Note that

$$[\mathbb{F}_{\ell}(\zeta_{2^{n}}):\mathbb{F}_{\ell}]=2^{n-2}$$

for $\ell\equiv\pm 3\text{ mod }8$. The corresponding eigenvalue is

	$\displaystyle a_{0}+a_{1}\zeta+\ldots+a_{2^{n-2}-1}\zeta^{2^{n-2}-1}-a_{2^{n-2}-1}\zeta^{2^{n-2}+1}-\ldots-a_{1}\zeta^{2^{n-1}-1}$
	$\displaystyle=\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=1}^{2^{n-2}-1}a_{2^{n-2}-j}\zeta^{2^{n-2}+j}.$

We want to show that the eigenvalues are all non-zero unless the matrix identically zero.

By Lemma 4.5, $\zeta^{2^{n-2}}\equiv\pm\sqrt{-2}\zeta^{2^{n-3}}+1$. Without loss of generality, we can assume

$$\zeta^{2^{n-2}}\equiv\sqrt{-2}\zeta^{2^{n-3}}+1$$

by choosing the sign of $\sqrt{-2}$.

Then

$$\zeta^{2^{n-2}+2^{n-3}}\equiv\sqrt{-2}\zeta^{2^{n-2}}+\zeta^{2^{n-3}}\equiv-\zeta^{2^{n-3}}+\sqrt{-2}.$$

We can use this to reduce the eigenvalue to lower terms.

	$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=1}^{2^{n-2}-1}a_{2^{n-2}-j}\zeta^{2^{n-2}+j}$
	$\displaystyle=a_{0}+a_{2^{n-3}}\zeta^{2^{n-3}}+\sum_{j=1}^{2^{n-3}-1}(a_{j}\zeta^{j}+a_{2^{n-3}+j}\zeta^{2^{n-3}+j})$
	$\displaystyle\hskip 7.22743pt-a_{2^{n-3}}\zeta^{2^{n-2}+2^{n-3}}-\sum_{j=1}^{2^{n-3}-1}(a_{2^{n-2}-j}\zeta^{2^{n-2}+j}+a_{2^{n-3}-j}\zeta^{2^{n-2}+2^{n-3}+j})$
	$\displaystyle=a_{0}-\sqrt{-2}a_{2^{n-3}}+2a_{2^{n-3}}\zeta^{2^{n-3}}+\sum_{j=1}^{2^{n-3}-1}\big{(}(a_{j}-a_{2^{n-2}-j}-\sqrt{-2}a_{2^{n-3}-j})\zeta^{j}\big{)}$
	$\displaystyle\hskip 14.45377pt+\sum_{j=1}^{2^{n-3}-1}\big{(}(a_{2^{n-3}+j}-\sqrt{-2}a_{2^{n-2}-j}+a_{2^{n-3}-j})\zeta^{2^{n-3}+j}\big{)}$

The eigenvalue is a polynomial in $\zeta$ of degree $2^{n-2}-1$. Therefore we can write the eigenvalue as a $2^{n-2}\times 2^{n-2}$ matrix, which we’ll call $B$, multiplied by the column vector $(a_{0},a_{1},\ldots,a_{2^{n-2}-1})$. Row $j$ corresponds to the coefficient of $\zeta^{j}$. We write $I$ to represent the identity matrix and $J$ to represent the anti-identity matrix. The anti-identity matrix is all zero except for ones on the diagonal that goes from the upper right corner to the lower left corner.

Then

$$B=\left(\begin{array}[]{c|c|c|c}1&&-\sqrt{-2}&\\ \hline\cr&I-\sqrt{-2}J&&-J\\ \hline\cr 0&&2&\\ \hline\cr&J&&I-\sqrt{-2}J\end{array}\right)$$

where the $I$ and $J$ matrices are both of size $(2^{n-3}-1)\times(2^{n-3}-1)$. The eigenvalue is zero if and only if the product of $B$ and the column vector $(a_{0},a_{1},\ldots,a_{2^{n-2}-1})$ is zero.

Lemma 4.6.

The matrix

$$\left(\begin{array}[]{c|c}I-\sqrt{-2}J&-J\\ \hline\cr J&I-\sqrt{-2}J\end{array}\right)$$

has full rank.

Proof.

We have the following chain of row operations:

	$\displaystyle\left(\begin{array}[]{c|c}I-\sqrt{-2}J&-J\\ \hline\cr J&I-\sqrt{-2}J\end{array}\right)$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I-\sqrt{-2}J&-J\\ \hline\cr\sqrt{-2}J&\sqrt{-2}I+2J\end{array}\right)\hskip 21.68121pt\text{(multiply 2nd row by }\sqrt{-2}\text{)}$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&\sqrt{-2}I+J\\ \hline\cr\sqrt{-2}J&\sqrt{-2}I+2J\end{array}\right)\hskip 21.68121pt\text{(add 2nd row to 1st)}$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&\sqrt{-2}I+J\\ \hline\cr I&J-\sqrt{-2}I\end{array}\right)\hskip 21.68121pt\text{(multiply 2nd row by }\frac{1}{\sqrt{-2}}J\text{)}$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&I+J\\ \hline\cr 0&-2\sqrt{-2}I\end{array}\right)\hskip 21.68121pt\text{(2nd row - 1st row }\text{)}$

This matrix has non-zero determinant, and therefore $B$ has non-zero determinant. ∎

Therefore $B$ has full rank, and so there are no non-trivial solutions to

$$B\cdot(a_{0},\ldots,a_{2^{n-2}-1})^{T}=(0,\ldots,0).$$

Therefore the eigenvalue equals zero only if $a_{j}=0$ for all $j$. So either the matrix of norm residue symbols (Equation 4.1) is identically zero or it has full rank.

$\bullet\hskip 10.00002ptp\equiv 3\text{ mod }8$, $\ell\equiv 3\text{ mod }8$:

Since $p\equiv 3\text{ mod }8$, no primes lying over $p$ are fixed by $\tau$. Without loss of generality, we can choose $\mathfrak{p}$ so that $\tau\mathfrak{p}=\sigma\mathfrak{p}$. Then

$$\tau\sigma^{j}\mathfrak{p}=\sigma^{-j}\tau\mathfrak{p}=\sigma^{-j+1}\mathfrak{p}.$$

Again construct a matrix $a_{ij}$ over $\mathbb{F}_{\ell}$ using the norm residue symbols:

$$\left(\frac{u_{j}}{\mathfrak{p}_{i}}\right)\equiv\omega^{a_{ij}}\text{ mod }\mathfrak{p}_{i}$$

where $\omega$ is a fixed primitive $\ell^{th}$ root of unity.

We’ll again start by considering $\mathfrak{p}_{0}=\mathfrak{p}$. Write

$$\left(\frac{u_{j}}{\mathfrak{p}}\right)=\omega^{a_{j}}.$$

Then using $\tau u_{j}=u_{2^{n-1}-j}^{-1}$ and the assumption that $\tau\mathfrak{p}=\sigma\mathfrak{p}$, for $0\leq j<2^{n-2}$,

$$\omega^{a_{j}}=\left(\frac{u_{j}}{\mathfrak{p}}\right)=\tau\left(\frac{u_{j}}{\mathfrak{p}}\right)=\left(\frac{u_{2^{n-1}-j}}{\sigma\mathfrak{p}}\right)^{-1}.$$

Apply $\sigma^{2^{n}-1}$:

$$\omega^{a_{j}}=\sigma^{2^{n}-1}(\omega^{a_{j}})=\left(\frac{u^{-1}_{2^{n-1}-j-1}}{\mathfrak{p}}\right)=\omega^{-a_{2^{n-1}-j-1}}.$$

Therefore the first row of the matrix is

$$\begin{pmatrix}a_{0}&a_{1}&\ldots&a_{2^{n-2}-1}&-a_{2^{n-2}-1}&\ldots&-a_{0}\end{pmatrix}.$$

We can again apply $\sigma$ to get the full matrix:

(4.2)

$$\begin{pmatrix}a_{0}&a_{1}&\ldots&a_{2^{n-2}-1}&-a_{2^{n-2}-1}&\ldots&-a_{0}\\ a_{0}&a_{0}&\ldots&a_{2^{n-2}-2}&a_{2^{n-2}-1}&\ldots&-a_{1}\\ a_{1}&a_{0}&\ldots&a_{2^{n-2}-3}&a_{2^{n-2}-2}&\ldots&-a_{2}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\ a_{2}&a_{3}&\ldots&-a_{2^{n-2}-2}&-a_{2^{n-2}-3}&\ldots&-a_{1}\\ a_{1}&a_{2}&\ldots&-a_{2^{n-2}-1}&-a_{2^{n-2}-2}&\ldots&-a_{0}\\ \end{pmatrix}$$

If $n=2$, we have the matrix

$$\begin{pmatrix}a_{0}&-a_{0}\\ a_{0}&a_{0}\\ -a_{0}&a_{0}\\ -a_{0}&-a_{0}\end{pmatrix}$$

which has rank 0 if $a_{0}=0$ and rank $2$ if $a_{0}\neq 0$.

For $n>2$, we again apply Proposition 4.4. The top half of the matrix has eigenvectors

$$(1,\zeta,\ldots,\zeta^{2^{n-1}-1})$$

where $\zeta$ is one of the $2^{n-1}$ distinct solutions to $x^{2^{n-1}}+1=0$. The corresponding eigenvalue is

	$\displaystyle a_{0}+a_{1}\zeta+\ldots+a_{2^{n-2}-1}\zeta^{2^{n-2}-1}-a_{2^{n-2}-1}\zeta^{2^{n-2}}-\ldots-a_{0}\zeta^{2^{n-1}-1}$
	$\displaystyle=\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=0}^{2^{n-2}-1}a_{2^{n-2}-1-j}\zeta^{2^{n-2}+j}.$

For $n>2$, we again have

	$\displaystyle x^{2^{n-1}}+1$	$\displaystyle\equiv(x^{2^{n-2}}+\sqrt{-2}x^{2^{n-3}}-1)(x^{2^{n-2}}-\sqrt{-2}x^{2^{n-3}}-1),$
	$\displaystyle\zeta^{2^{n-2}}$	$\displaystyle\equiv\sqrt{-2}\zeta^{2^{n-3}}+1,$
	$\displaystyle\zeta^{2^{n-2}+2^{n-3}}$	$\displaystyle\equiv-\zeta^{2^{n-3}}+\sqrt{-2}.$

We can use this to reduce the eigenvalue to lower terms.

	$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=0}^{2^{n-2}-1}a_{2^{n-2}-1-j}\zeta^{2^{n-2}+j}$
	$\displaystyle=\sum_{j=0}^{2^{n-3}-1}\left(a_{j}\zeta^{j}+a_{2^{n-3}+j}\zeta^{2^{n-3}+j}-a_{2^{n-2}-1-j}\zeta^{2^{n-2}+j}-a_{2^{n-3}-1-j}\zeta^{2^{n-2}+2^{n-3}+j}\right)$
	$\displaystyle\equiv\sum_{j=0}^{2^{n-3}-1}\big{(}(a_{j}-a_{2^{n-2}-1-j}-\sqrt{-2}a_{2^{n-3}-1-j})\zeta^{j}$
	$\displaystyle\hskip 7.22743pt+(a_{2^{n-3}+j}-\sqrt{-2}a_{2^{n-2}-1-j}+a_{2^{n-3}-1-j})\zeta^{2^{n-3}+j}\big{)}$

Writing the matrix $B$ corresponding to the eigenvector as in the previous case, we obtain

$$B=\left(\begin{array}[]{c|c}I-\sqrt{-2}J&-J\\ \hline\cr J&I-\sqrt{-2}J\end{array}\right)$$

where the $I$ and $J$ matrices are both of size $(2^{n-3})\times(2^{n-3})$. So by Lemma 4.6, the matrix has full rank and so there are no non-trivial solutions for the $\{a_{j}\}$ that yield a zero eigenvalue.

$\bullet\hskip 10.00002ptp\equiv 7\text{ mod }8$, $\ell\equiv 5\text{ mod }8$:

In this case, we again have a fixed prime: $\tau\mathfrak{p}=\mathfrak{p}$. Then

$$\tau\sigma^{j}\mathfrak{p}=\sigma^{-j}\tau\mathfrak{p}=\sigma^{-j}\mathfrak{p}.$$

We then construct the matrix $(a_{ij})$ using the norm residue symbols to get the same skew circulant matrix as in Equation 4.1.

The eigenvalues of the matrix are of the form

$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=1}^{2^{n-2}-1}a_{2^{n-2}-j}\zeta^{2^{n-2}+j}.$

for $\zeta$ a solution to $x^{2^{n-1}}+1=0$. We want to show that the eigenvalues are all non-zero unless the matrix is the zero matrix.

The case $n=2$ is again trivial. For $n>2$, we use Lemma 4.5, which tells us that $\zeta^{2^{n-2}}\equiv\pm\sqrt{-1}$. Again, we assume without loss of generality that $\zeta^{2^{n-2}}\equiv\sqrt{-1}$.

We can use this to reduce the eigenvalue to lower terms.

		$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=1}^{2^{n-2}-1}a_{2^{n-2}-j}\zeta^{2^{n-2}+j}$
	$\displaystyle\equiv$	$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sqrt{-1}\sum_{j=1}^{2^{n-2}-1}a_{2^{n-2}-j}\zeta^{j}$
	$\displaystyle\equiv$	$\displaystyle\hskip 7.22743pta_{0}+\sum_{j=1}^{2^{n-2}-1}(a_{j}-\sqrt{-1}a_{2^{n-2}-j})\zeta^{j}$

We write the eigenvalue as a $2^{n-2}\times 2^{n-2}$ matrix $B$ multiplied by the column vector $(a_{0},a_{1},\ldots,a_{2^{n-2}-1})$ just as in the previous cases.

Then

$$B=\left(\begin{array}[]{c|c|c|c}1&&0&\\ \hline\cr&I&&-\sqrt{-1}J\\ \hline\cr 0&&1-\sqrt{-1}&\\ \hline\cr&-\sqrt{-1}J&&I\\ \end{array}\right)$$

where the $I$ and $J$ matrices are both of dimension $(2^{n-3}-1)\times(2^{n-3}-1)$.

Lemma 4.7.

The matrix

$$\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr-\sqrt{-1}J&I\end{array}\right)$$

has full rank.

Proof.

We have the following chain of row operations:

	$\displaystyle\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr-\sqrt{-1}J&I\end{array}\right)$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr-I&-\sqrt{-1}J\end{array}\right)\hskip 21.68121pt\text{(multiply 2nd row by }-\sqrt{-1}J\text{)}$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr 0&-2\sqrt{-1}J\end{array}\right)\hskip 21.68121pt\text{(add 1st row to 2nd)}$
	$\displaystyle\rightarrow\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr 0&2I\end{array}\right)\hskip 21.68121pt\text{(multiply 2nd row by }\sqrt{-1}J\text{)}$

This matrix has non-zero determinant. ∎

Therefore $B$ has full rank and so the only way for the eigenvalue to equal zero is if $a_{j}=0$ for all $j$.

$\bullet\hskip 10.00002ptp\equiv 3\text{ mod }8$, $\ell\equiv 5\text{ mod }8$:

Here, the matrix $(a_{ij})$ derived from the norm residue symbols is of the same form as in Equation 4.2. Its eigenvalues are of the form

$\displaystyle\sum_{j=0}^{2^{n-2}-1}a_{j}\zeta^{j}-\sum_{j=0}^{2^{n-2}-1}a_{2^{n-2}-1-j}\zeta^{2^{n-2}+j}.$

By Lemma 4.5, $\zeta^{2^{n-2}}\equiv\sqrt{-1}$. Therefore the eigenvalue can be reduced to

$\displaystyle\sum_{j=0}^{2^{n-2}-1}(a_{j}-\sqrt{-1}a_{2^{n-2}-1-j})\zeta^{j}.$

This eigenvalue, when embedded into a matrix $B$, is of the form

$$\left(\begin{array}[]{c|c}I&-\sqrt{-1}J\\ \hline\cr-\sqrt{-1}J&I\end{array}\right)$$

where $I$ and $J$ have dimension $2^{n-3}\times 2^{n-3}$. By Lemma 4.7, we know this matrix has non-zero determinant.