Let $f(x)=x^{101}-x$ and $g(x)=x^{101}-x+1$ in ring $\mathbb{Z}_{101}[x]$.
1) Are quotient rings $$\mathbb{Z}_{101}[x]/< f(x) > and ~~\mathbb{Z}_{101}[x]/<g(x)>$$ isomorphic?
2) How many of the invertible elements of $\mathbb{Z}_{101}[x]/ <f(x)> $ are equal to their inverse?
Attempt. 1) I believe not. Since $f(x)=x(x-1)\ldots (x-100)$ in $\mathbb{Z}_{101}[x]$, quotient ring $\mathbb{Z}_{101}[x]/ <f(x)>$, having $101^{100}$ elements, in not an integral domain. On the other hand, quotient ring $\mathbb{Z}_{101}[x]/<g(x)>$ has the same number of elements, i.e. $101^{100}$, but I am having difficulty to prove that it is not an integral domain ($g(x)$ has no roots in $\mathbb{Z}_{101}$, but I am not sure if this helps).
2) The number of invertible elements of $\mathbb{Z}_{101}[x]/ <f(x)>$ is: \begin{eqnarray}&&101^{101}-\big(100\cdot 101+100\cdot 101^2+\ldots+100\cdot 101^{100}\big)-1\nonumber\\ &=&101^{101}-100\cdot 101\cdot \big(1+101+\ldots+101^{99}\big)-1\nonumber\\ &=&101^{101}-100\cdot 101 \cdot \frac{101^{100}-1}{101-1}-1\nonumber\\ &=&101^{101}-101 \cdot (101^{100}-1)-1=101-1=100.\nonumber \end{eqnarray} In order for an invertible element $$h(x)+<f(x)>=a_{100}x^{100}+\ldots+a_0+<f(x)>$$ of $\mathbb{Z}_{101}[x]/ <f(x)>$ to be equal to its inverse, we must have: $$h(x)+<f(x)>=(h(x)+<f(x)>)^{-1} \iff$$ $$f(x)~|~(h(x))^2-1=\big(h(x)-1\big)\big(h(x)+1\big)\iff$$ $$\forall k\in \mathbb{Z}_{101}:~x-k~|~h(x)-1 ~~or ~~h(x)+1.$$ But how would this give the necessary conditions on the coefficients of $h(x)$, to determine the exact number of the desired elements?
Thanks in advance.
Note that $101$ is prime, so that $\mathbb{Z}_{101}$ is a field; neither polynomial has repeated root (in its splitting field) because the derivative polynomial is in both cases $-1$.
If we have a monic polynomial $h(x)=h_1(x)h_2(x)\dots h_k(x)$ over a field $F$, with $h_1,h_2,\dots,h_n$ monic irreducible and pairwise distinct, the Chinese remainder theorem tells us that $$ F[x]/\langle h(x)\rangle\cong F[x]/\langle h_1(x)\rangle\times F[x]/\langle h_2(x)\rangle\times\dots\times F[x]/\langle h_k(x)\rangle \tag{*} $$ where each factor is a field extension of $F$.
By Fermat's little theorem, $f(x)=x^{101}-x$ factors over $\mathbb{Z}_{101}$ as a product of linear factors, so $$ \mathbb{Z}_{101}[x]/\langle f(x)\rangle\cong(\mathbb{Z}_{101})^{101}\tag{**} $$ Since $g(x)=x^{101}-x+1$ has no roots in $\mathbb{Z}_{101}$, but can nonetheless be factored as $g=g_1g_2\dots g_k$, with $g_1,\dots,g_k$ monic irreducible and $k<101$, we have that $\mathbb{Z}_{101}[x]/\langle g(x)\rangle$ is a product of $k$ fields. Since $k\ne101$, we can conclude that the two rings are not isomorphic (for instance, show that the number of elements equal to their inverse is not the same).
For part (b), the sought number is $2^{101}$ as follows from the isomorphism (**).
Note. As it has been remarked in comments, $g(x)$ is actually irreducible, so $k=1$ and $\mathbb{Z}_{101}[x]/\langle g(x)\rangle$ is a field. However this is not needed for the present exercise. What's needed is that the number of irreducible factors of $g$ is less than $101$.