Let $C$ be smooth projective curve, $\mathcal{M}$ be the moduli space of semistable Higgs bundles, $H:\mathcal{M}\rightarrow W= \bigoplus H^0(C,K_{C}^{\otimes i})$ be the Hitchin map, and $\pi :C_s\rightarrow C$ be the smooth spectral curve associated to $s=(s_1,...,s_n)\in W$.

I'm trying to show that $\{s\in W\mid C_s$ is irreducible and reduced.$\}$ and $\{s\in W\mid C_s$ is smooth.$\}$ are both open in $W$.
It is in the section 3 of this paper of A. Beauville, M.S. Narasimhan, and S. Ramanan.

$C_s$ is a divisor in $\mathbb{V}(K_C^{\otimes r})$ corresponding to $x^r+s_1\cdot x+\cdots+s_n$. So, locally the definition equation is a polynomial.
Let $(z,U)$ be a local coodinate of $C$ near $z$. For smoothness, we can use Jacobian criterion. But, is this right ?

For irreducibility, is $C_s$ irreducible iff the corresponding definition equation is irreducible for every $z\in C$ ?
I don't think it's right because this is just a local condition. And I'm stack.

I think this is a very easy and dumb question. Thanks in advance.

Let $C$ be smooth projective curve, $\mathcal{M}$ be the moduli space of semistable Higgs bundles, $H:\mathcal{M}\rightarrow W= \bigoplus H^0(C,K_{C}^{\otimes i})$ be the Hitchin map, and $\pi :C_s\rightarrow C$ be the smooth spectral curve associated to $s=(s_1,...,s_n)\in W$.

I'm trying to show that $\{s\in W\mid C_s$ is irreducible and reduced.$\}$ and $\{s\in W\mid C_s$ is smooth.$\}$ are both open in $W$.
It is in the section 3 of this paper of A. Beauville, M.S. Narasimhan, and S. Ramanan.

$C_s$ is a divisor in $\mathbb{V}(K_C)$ corresponding to $x^r+s_1\cdot x+\cdots+s_n$. So, locally the definition equation is a polynomial.
Let $(z,U)$ be a local coodinate of $C$ near $z$. For smoothness, we can use Jacobian criterion. But, is this right ?

For irreducibility, is $C_s$ irreducible iff the corresponding definition equation is irreducible for every $z\in C$ ?
I don't think it's right because this is just a local condition. And I'm stack.

I think this is a very easy and dumb question. Thanks in advance.