In this section, we will consider the freeness for hypersurface arrangements containing hyperplanes and spheres, and give the explicit bases for the derivation modules of the free ones. First, we show that the hypersurface arrangement having *n* spheres is free.

#### Theorem 3.1

*Let* 𝓜_{n} *be the hypersurface arrangement consisting of* *n spheres* 𝓢_{1},…,𝓢_{n}, *where*

$$\begin{array}{}{\mathcal{S}}_{i}=\{({x}_{1},\dots ,{x}_{\ell})\mid \sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}={r}_{i},\text{\hspace{0.17em}}({a}_{1},{a}_{2},\dots ,{a}_{\ell})\in {\mathbb{R}}^{\ell},\text{\hspace{0.17em}}{r}_{i}\in {\mathbb{R}}^{+}\}.\end{array}$$

*Define derivations*
$\begin{array}{}{\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n}\end{array}$ *by*

$$\begin{array}{}\mathsf{M}({\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n})={A}_{n}{A}_{n-1}\cdots {A}_{1},\end{array}$$

*where A*_{i} is an *ℓ* × *ℓ* *matrix and the* (*p*, *q*) *entry of A*_{i} is

$$\begin{array}{}({A}_{i}{)}_{pq}=\left\{\begin{array}{ll}({x}_{p}-{a}_{p})({x}_{q}-{a}_{q})& \text{\hspace{0.17em}}if\text{\hspace{0.17em}}p\ne q,\\ ({x}_{p}-{a}_{p}{)}^{2}-{r}_{i}& \text{\hspace{0.17em}}if\text{\hspace{0.17em}}p=q.\end{array}\right.\end{array}$$

*Then*
$\begin{array}{}{\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n}\end{array}$ *form a basis for D*(𝓜_{n}) *and* exp 𝓜_{n} = {2*n*,…,2*n*}, *where* 2*n appears* *ℓ times*.

#### Proof

We will prove this result by Theorem 2.9: Saito’s criterion. By Lemma 2.7, we obtain

$$\begin{array}{}det{A}_{i}=(-{r}_{i}{)}^{\ell -1}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{i}].\end{array}$$

Therefore,

$$\begin{array}{}det\mathsf{M}({\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n})=\prod _{i=1}^{n}det{A}_{i}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\doteq \prod _{i=1}^{n}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{i}]\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=Q({\mathcal{M}}_{n}).\end{array}$$

Next, we will prove
$\begin{array}{}{\phi}_{i}^{n}\end{array}$ ∈ *D*(𝓜_{n}) and deg
$\begin{array}{}{\phi}_{i}^{n}\end{array}$ = 2*n* for any 1 ≤ *i* ≤ *ℓ* by induction on *n*.

The case *n* = 1 is clear according to Lemma 2.8 and Corollary 2.10. For the case *n* + 1, we notice that

$$\begin{array}{}{\phi}_{i}^{n+1}=\sum _{p=1}^{\ell}{\phi}_{i}^{n+1}({x}_{p}){D}_{p}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[\sum _{q=1}^{\ell}({A}_{n+1}{)}_{pq}{\phi}_{i}^{n}({x}_{q})]{D}_{p}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[\sum _{q\ne p}({x}_{p}-{a}_{p})({x}_{q}-{a}_{q}){\phi}_{i}^{n}({x}_{q})+[({x}_{p}-{a}_{p}{)}^{2}-{r}_{n+1}]{\phi}_{i}^{n}({x}_{p})]{D}_{p}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[({x}_{p}-{a}_{p})\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}){\phi}_{i}^{n}({x}_{q})]{D}_{p}-{r}_{n+1}\sum _{p=1}^{\ell}{\phi}_{i}^{n}({x}_{p}){D}_{p}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}){\phi}_{i}^{n}({x}_{q})\sum _{p=1}^{\ell}({x}_{p}-{a}_{p}){D}_{p}-{r}_{n+1}{\phi}_{i}^{n}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{1}{2}{\phi}_{i}^{n}[\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}{)}^{2}]\sum _{p=1}^{\ell}({x}_{p}-{a}_{p}){D}_{p}-{r}_{n+1}{\phi}_{i}^{n}.\end{array}$$

Therefore, for 1 ≤ *i* ≤ *ℓ* and 1 ≤ *j* ≤ *n*,

$$\begin{array}{}& {\phi}_{i}^{n+1}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{j}]\\ & ={\phi}_{i}^{n}[\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}{)}^{2}]\sum _{p=1}^{\ell}({x}_{p}-{a}_{p}{)}^{2}-{r}_{n+1}{\phi}_{i}^{n}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{j}]\\ & ={\phi}_{i}^{n}[\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}{)}^{2}][\sum _{p=1}^{\ell}({x}_{p}-{a}_{p}{)}^{2}-{r}_{n+1}],\end{array}$$

by induction hypothesis,

$$\begin{array}{}{\phi}_{i}^{n}[\sum _{q=1}^{\ell}({x}_{q}-{a}_{q}{)}^{2}]\in \prod _{j=1}^{n}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{j}]S,\end{array}$$

hence,

$$\begin{array}{}{\phi}_{i}^{n+1}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{j}]\in \prod _{j=1}^{n+1}[\sum _{k=1}^{\ell}({x}_{k}-{a}_{k}{)}^{2}-{r}_{j}]S.\end{array}$$

This means
$\begin{array}{}{\phi}_{i}^{n+1}\in D({\mathcal{M}}_{n+1})\end{array}$ for any 1 ≤ *i* ≤ *ℓ*, in addition,

$$\begin{array}{}\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\phi}_{i}^{n+1}=\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\phi}_{i}^{n}+2=2n+2=2(n+1).\end{array}$$

We complete the induction, so by Saito’s criterion
$\begin{array}{}{\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n}\end{array}$ form a basis for 𝓜_{n}, and exp 𝓜_{n} = {2*n*,…,2*n*}. □

Next, we will study the freeness for the hypersurface arrangements consisting of hyperplanes and spheres, where all the spheres are centered at origin.

#### Theorem 3.2

*Assume* 𝓐 *is a free hyperplane arrangement with a homogeneous basis* *θ*_{1},…,*θ*_{ℓ}, *and* exp 𝓐 = {*d*_{1},…,*d*_{ℓ}},
$\begin{array}{}{\mathcal{S}}_{i}^{0}\end{array}$ *is the sphere centered at origin*:

$$\begin{array}{}{\mathcal{S}}_{i}^{0}=\{({x}_{1},\dots ,{x}_{\ell})\mid \sum _{k=1}^{\ell}{x}_{k}^{2}={r}_{i}\in {\mathbb{R}}^{+}\},1\le i\le n,\end{array}$$

*and*

$$\begin{array}{}{\mathcal{M}}_{n}=\mathcal{A}\cup \{{\mathcal{S}}_{1}^{0},\dots ,{\mathcal{S}}_{n}^{0}\},\end{array}$$

*Define derivations*
$\begin{array}{}{\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n}\end{array}$ *by*

$$\begin{array}{}\mathsf{M}({\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n})=({A}_{n}{A}_{n-1}\cdots {A}_{1})\mathsf{M}({\theta}_{1},\dots ,{\theta}_{\ell}),\end{array}$$

*where A*_{i} is an *ℓ* × *ℓ* *matrix and the* (*p*, *q*) *entry of A*_{i} is

$$\begin{array}{}({A}_{i}{)}_{pq}=\left\{\begin{array}{ll}{x}_{p}{x}_{q}& \text{\hspace{0.17em}}if\text{\hspace{0.17em}}p\ne q,\\ {x}_{p}^{2}-{r}_{i}& \text{\hspace{0.17em}}if\text{\hspace{0.17em}}p=q,\end{array}\right.\end{array}$$

*then*
$\begin{array}{}{\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n}\end{array}$ *form a basis for D*(𝓜_{n}) *and* exp 𝓜_{n} = {*d*_{1} + 2*n*,…,*d*_{ℓ} + 2*n*}.

#### Proof

By Lemma 2.7, we obtain

$$\begin{array}{}det{A}_{i}=(-{r}_{i}{)}^{\ell -1}(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{i}).\end{array}$$

Since 𝓐 is a free arrangement with a homogeneous basis *θ*_{1},…,*θ*_{ℓ}, by Saito’s criterion,

$$\begin{array}{}det\mathsf{M}({\theta}_{1},\dots ,{\theta}_{\ell})\doteq Q(\mathcal{A}).\end{array}$$

Therefore,

$$\begin{array}{}det\mathsf{M}({\phi}_{1}^{n},\dots ,{\phi}_{\ell}^{n})=\prod _{i=1}^{n}(det{A}_{i})det\mathsf{M}({\theta}_{1},\dots ,{\theta}_{\ell})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\doteq \prod _{i=1}^{n}(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{i})Q(\mathcal{A})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=Q({\mathcal{M}}_{n}).\end{array}$$

Next, we will prove
$\begin{array}{}{\phi}_{i}^{n}\end{array}$ ∈ *D*(𝓜_{n}) and deg
$\begin{array}{}{\phi}_{i}^{n}\end{array}$ = *d*_{i} + 2*n* for any 1 ≤ *i* ≤ *ℓ* by induction on *n*.

For the case *n* = 1,

$$\begin{array}{}{\phi}_{i}^{1}=\sum _{p=1}^{\ell}{\phi}_{i}^{1}({x}_{p}){D}_{p}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[\sum _{q=1}^{\ell}({A}_{1}{)}_{pq}{\theta}_{i}({x}_{q})]{D}_{p}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[\sum _{q\ne p}{x}_{p}{x}_{q}{\theta}_{i}({x}_{q})+({x}_{p}^{2}-{r}_{1}){\theta}_{i}({x}_{p})]{D}_{p}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{p=1}^{\ell}[{x}_{p}\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q})]{D}_{p}-{r}_{1}\sum _{p=1}^{\ell}{\theta}_{i}({x}_{p}){D}_{p}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q})\sum _{p=1}^{\ell}{x}_{p}{D}_{p}-{r}_{1}{\theta}_{i}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q}){\theta}_{E}-{r}_{1}{\theta}_{i}.\end{array}$$

Since *θ*_{E}, *θ*_{i} ∈ *D*(𝓐), we have
$\begin{array}{}{\phi}_{i}^{1}\end{array}$ ∈ *D*(𝓐) for any 1 ≤ *i* ≤ *ℓ*. And

$$\begin{array}{}\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\phi}_{i}^{1}=\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}[\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q}){\theta}_{E}]=\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\theta}_{i}+2={d}_{i}+2.\end{array}$$

In addition,

$$\begin{array}{}{\phi}_{i}^{1}(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{1})=[\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q}){\theta}_{E}-{r}_{1}{\theta}_{i}](\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{1})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q})(2\sum _{k=1}^{\ell}{x}_{k}^{2})-2{r}_{1}\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=2(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{1})\sum _{q=1}^{\ell}{x}_{q}{\theta}_{i}({x}_{q})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\in (\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{1})S\end{array}$$

that is,
$\begin{array}{}{\phi}_{i}^{1}\in D(\{{\mathcal{S}}_{1}^{0}\}).\end{array}$ Therefore,
$\begin{array}{}{\phi}_{i}^{1}\in D(\mathcal{A})\cap D(\{{\mathcal{S}}_{1}^{0}\})=D({\mathcal{M}}_{1})\end{array}$ for any 1 ≤ *i* ≤ *ℓ*.

For the case *n* + 1, by the similar calculation of
$\begin{array}{}{\phi}_{i}^{1}\end{array}$, we get

$$\begin{array}{}{\phi}_{i}^{n+1}=\sum _{q=1}^{\ell}{x}_{q}{\phi}_{i}^{n}({x}_{q}){\theta}_{E}-{r}_{n+1}{\phi}_{i}^{n}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{1}{2}{\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2}){\theta}_{E}-{r}_{n+1}{\phi}_{i}^{n}.\end{array}$$

By induction hypothesis,

$$\begin{array}{}{\phi}_{i}^{n}\in D({\mathcal{M}}_{n})\subseteq D(\bigcup _{i=1}^{n}\{{\mathcal{S}}_{i}^{0}\}),\end{array}$$

we obtain

$$\begin{array}{}{\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2})\in \prod _{j=1}^{n}(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{j})S.\end{array}$$

Therefore,

$$\begin{array}{}{\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2}){\theta}_{E}\in D(\bigcup _{i=1}^{n}\{{\mathcal{S}}_{i}^{0}\}).\end{array}$$

Combining *θ*_{E} ∈ *D*(𝓐) with

$$\begin{array}{}D({\mathcal{M}}_{n})=D(\bigcup _{i=1}^{n}\{{\mathcal{S}}_{i}^{0}\})\bigcap D(\mathcal{A}),\end{array}$$

we conclude

$$\begin{array}{}{\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2}){\theta}_{E}\in D({\mathcal{M}}_{n}).\end{array}$$

Hence,
$\begin{array}{}{\phi}_{i}^{n+1}\in D({\mathcal{M}}_{n})\end{array}$ since
$\begin{array}{}{\phi}_{i}^{n}\in D({\mathcal{M}}_{n}).\end{array}$

In addition, we have

$$\begin{array}{}& {\phi}_{i}^{n+1}(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{n+1})\\ & ={\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2})\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{n+1}{\phi}_{i}^{n}(\sum _{k=1}^{\ell}{x}_{k}^{2})\\ & =(\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{n+1}){\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2})\\ & \in (\sum _{k=1}^{\ell}{x}_{k}^{2}-{r}_{n+1})S,\end{array}$$

We obtain
$\begin{array}{}{\phi}_{i}^{n+1}\in D(\{{\mathcal{S}}_{n+1}^{0}\})\end{array}$ for any 1 ≤ *i* ≤ *ℓ*, therefore

$$\begin{array}{}{\phi}_{i}^{n+1}\in D(\{{\mathcal{S}}_{n+1}^{0}\})\cap D({\mathcal{M}}_{n})=D({\mathcal{M}}_{n+1}),\text{\hspace{0.17em}}1\le i\le \ell .\end{array}$$

Moreover,

$$\begin{array}{}\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\phi}_{i}^{n+1}=\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}[{\phi}_{i}^{n}(\sum _{q=1}^{\ell}{x}_{q}^{2}){\theta}_{E}]=\mathrm{p}\mathrm{d}\mathrm{e}\mathrm{g}{\phi}_{i}^{n}+2={d}_{i}+2n+2={d}_{i}+2(n+1),\text{\hspace{0.17em}}1\le i\le \ell .\end{array}$$

We complete the induction. □

#### Corollary 3.3

*Let* 𝓜_{n} =
$\begin{array}{}\mathcal{A}\cup \{{\mathcal{S}}_{1}^{0},\dots ,{\mathcal{S}}_{n}^{0}\}\end{array}$ *be the hypersurface arrangement defined in Theorem 3.2*. *Then* 𝓐 *is free if and only if* 𝓜_{n} *is free*.

#### Proof

If 𝓐 is free we can obtain that 𝓜_{n} is free directly from Theorem 3.2. Assume 𝓜_{n} is free, 𝓐 ⊆ 𝓜_{n}, then *D*(𝓜_{n}) ⊆ *D*(𝓐). Let *φ*_{1},…,*φ*_{ℓ} be a basis for *D*(𝓜_{n}), then *φ*_{i} ∈ *D*(𝓐) for 1 ≤ *i* ≤ *ℓ*. Write
$\begin{array}{}{\phi}_{i}=\sum _{k\ge 0}{\phi}_{i}^{(k)},\end{array}$ where
$\begin{array}{}{\phi}_{i}^{(k)}\end{array}$ is zero or homogeneous of degree *k* ≥ 0. Since *Q*(𝓐)*S* is generated by homogeneous polynomial *Q*(𝓐), each homogeneous component
$\begin{array}{}{\phi}_{i}^{(k)}(Q(\mathcal{A}))\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\phi}_{i}(Q(\mathcal{A}))\end{array}$ also lies in *Q*(𝓐)*S*. This shows that
$\begin{array}{}{\phi}_{i}^{(k)}\in D(\mathcal{A})\end{array}$ for *k* ≥ 0. Since
$\begin{array}{}[Q(\bigcup _{i=1}^{n}\{{\mathcal{S}}_{i}\})](0)\ne 0,\end{array}$ there exist
$\begin{array}{}{\phi}_{1}^{({d}_{1})},\dots ,{\phi}_{\ell}^{({d}_{\ell})}\end{array}$ such that

$$\begin{array}{}det\mathsf{M}({\phi}_{1}^{({d}_{1})},\dots ,{\phi}_{\ell}^{({d}_{\ell})})\doteq Q(\mathcal{A}).\end{array}$$

By Saito’s criterion,
$\begin{array}{}{\phi}_{1}^{({d}_{1})},\dots ,{\phi}_{\ell}^{({d}_{\ell})}\end{array}$ form a basis for *D*(𝓐). □

#### Example 3.5

*Let* 𝓜 *be a hypersurface arrangement with the defining polynomial*

$$\begin{array}{}Q(\mathcal{M})=({x}_{1}-{x}_{2})({x}_{1}-{x}_{3})({x}_{2}-{x}_{3})({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-1).\end{array}$$

*In this case*, *the hyperplane arrangement* 𝓐 ⊆ 𝓜 *is the Coxeter arrangement of type* *A*_{2}, *it is a free arrangement with* exp 𝓐 = {0, 1, 2}. *By Theorem 3.2*, 𝓜 *is a free hypersurface arrangement and* *D*(𝓜) *has the basis* *φ*_{1}, *φ*_{2}, *φ*_{3} *as follows*:

$$\begin{array}{}({\phi}_{1},{\phi}_{2},{\phi}_{3})=({D}_{1},{D}_{2},{D}_{3})\left(\begin{array}{ccc}{x}_{1}^{2}-1& {x}_{1}{x}_{2}& {x}_{1}{x}_{3}\\ {x}_{2}{x}_{1}& {x}_{2}^{2}-1& {x}_{2}{x}_{3}\\ {x}_{3}{x}_{1}& {x}_{3}{x}_{2}& {x}_{3}^{2}-1\end{array}\right)\left(\begin{array}{ccc}1& {x}_{1}& {x}_{1}^{2}\\ 1& {x}_{2}& {x}_{2}^{2}\\ 1& {x}_{3}& {x}_{3}^{2}\end{array}\right).\end{array}$$

*That is*

$$\begin{array}{}{\phi}_{1}=({x}_{1}^{2}+{x}_{1}{x}_{2}+{x}_{1}{x}_{3}-1){D}_{1}+({x}_{1}{x}_{2}+{x}_{2}^{2}+{x}_{2}{x}_{3}-1){D}_{2}+({x}_{1}{x}_{3}+{x}_{2}{x}_{3}+{x}_{3}^{2}-1){D}_{3},\\ {\phi}_{2}={x}_{1}({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-1){D}_{1}+{x}_{2}({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-1){D}_{2}+{x}_{3}({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-1){D}_{3},\\ {\phi}_{3}={x}_{1}({x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}-{x}_{1}){D}_{1}+{x}_{2}({x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}-{x}_{2}){D}_{2}+{x}_{3}({x}_{1}^{3}+{x}_{2}^{3}+{x}_{3}^{3}-{x}_{3}){D}_{3}.\end{array}$$

*And* exp 𝓜 = {pdeg*φ*_{1}, pdeg*φ*_{2}, pdeg*φ*_{3}} = {2, 3, 4}.

#### Example 3.6

*Let* 𝓜 *be a hypersurface arrangement with the defining polynomial*

$$\begin{array}{}Q(\mathcal{M})={x}_{1}{x}_{2}{x}_{3}({x}_{1}+{x}_{2})({x}_{1}+{x}_{3})({x}_{2}+{x}_{3})({x}_{1}-{x}_{2})({x}_{1}-{x}_{3})({x}_{2}-{x}_{3})({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-1)({x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}-2).\end{array}$$

*In this case*, *the hyperplane arrangement* 𝓐 ⊆ 𝓜 *is the Coxeter arrangement of type* *B*_{3}, *it is a free arrangement with* exp 𝓐 = {1, 3, 5}. *By Theorem 3.2*, 𝓜 *is a free hypersurface arrangement and* *D*(𝓜) *has the basis* *φ*_{1}, *φ*_{2}, *φ*_{3} *as follows*:

$$\begin{array}{}({\phi}_{1},{\phi}_{2},{\phi}_{3})=({D}_{1},{D}_{2},{D}_{3})\left(\begin{array}{ccc}{x}_{1}^{2}-2& {x}_{1}{x}_{2}& {x}_{1}{x}_{3}\\ {x}_{2}{x}_{1}& {x}_{2}^{2}-2& {x}_{2}{x}_{3}\\ {x}_{3}{x}_{1}& {x}_{3}{x}_{2}& {x}_{3}^{2}-2\end{array}\right)\left(\begin{array}{ccc}{x}_{1}^{2}-1& {x}_{1}{x}_{2}& {x}_{1}{x}_{3}\\ {x}_{2}{x}_{1}& {x}_{2}^{2}-1& {x}_{2}{x}_{3}\\ {x}_{3}{x}_{1}& {x}_{3}{x}_{2}& {x}_{3}^{2}-1\end{array}\right)\left(\begin{array}{ccc}{x}_{1}& {x}_{1}^{3}& {x}_{1}^{5}\\ {x}_{2}& {x}_{2}^{3}& {x}_{2}^{5}\\ {x}_{3}& {x}_{3}^{3}& {x}_{3}^{5}\end{array}\right).\end{array}$$

*And* exp 𝓜 = {pdeg*φ*_{1}, pdeg*φ*_{2}, pdeg*φ*_{3}} = {5, 7, 9}.

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