Let $S$ be the set of $2\times2$-matrices over $\mathbb{F}_{p}$ with trace $1$ and determinant $0$. Determine $|S|$.
1. **Form of the Matrix:**
   Let \( S \) be the set of \( 2 \times 2 \) matrices over \( \mathbb{F}_p \) with trace 1 and determinant 0. A general \( 2 \times 2 \) matrix is given by:
   \[
   A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
   \]
   The trace condition \( \text{tr}(A) = 1 \) implies:
   \[
   a + d = 1
   \]
   The determinant condition \( \det(A) = 0 \) implies:
   \[
   ad - bc = 0
   \]
   Substituting \( d = 1 - a \) into the determinant condition, we get:
   \[
   a(1 - a) - bc = 0 \implies a(1 - a) = bc
   \]

2. **Case \( a = 0 \):**
   If \( a = 0 \), then the equation \( a(1 - a) = bc \) simplifies to \( 0 = bc \). This means either \( b = 0 \) or \( c = 0 \). Therefore, the possible matrices are:
   \[
   \begin{pmatrix} 0 & b \\ c & 1 \end{pmatrix} \quad \text{with} \quad b = 0 \text{ or } c = 0
   \]
   For each \( b \) (where \( b \) can take \( p \) values) and each \( c \) (where \( c \) can take \( p \) values), we have \( 2p - 1 \) possible matrices (since the pair \( (b, c) = (0, 0) \) is counted twice).

3. **Case \( a = 1 \):**
   If \( a = 1 \), then the equation \( a(1 - a) = bc \) simplifies to \( 0 = bc \). This means either \( b = 0 \) or \( c = 0 \). Therefore, the possible matrices are:
   \[
   \begin{pmatrix} 1 & b \\ c & 0 \end{pmatrix} \quad \text{with} \quad b = 0 \text{ or } c = 0
   \]
   Similarly, for each \( b \) and each \( c \), we have \( 2p - 1 \) possible matrices.

4. **Case \( a \neq 0, 1 \):**
   If \( a \neq 0 \) and \( a \neq 1 \), then \( a(1 - a) \neq 0 \). Hence, \( bc = a(1 - a) \neq 0 \). This implies \( b \neq 0 \) and \( c = \frac{a(1 - a)}{b} \). For each non-zero \( b \) (which can take \( p - 1 \) values), \( c \) is uniquely determined. Therefore, for each \( a \) (where \( a \) can take \( p - 2 \) values), we have \( p - 1 \) possible matrices.

5. **Counting the Total Number of Matrices:**
   - For \( a = 0 \), there are \( 2p - 1 \) matrices.
   - For \( a = 1 \), there are \( 2p - 1 \) matrices.
   - For \( a \neq 0, 1 \), there are \( (p - 2)(p - 1) \) matrices.

   Summing these, we get:
   \[
   |S| = (2p - 1) + (2p - 1) + (p - 2)(p - 1)
   \]
   Simplifying this expression:
   \[
   |S| = 2(2p - 1) + (p - 2)(p - 1) = 4p - 2 + p^2 - 3p + 2 = p^2 + p
   \]
   Therefore:
   \[
   |S| = p(p + 1)
   \]

The final answer is \( \boxed{p(p + 1)} \).