When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \minus{} 1$ the quotient is $ f(y)$ and the remainder is $ R_1$.  When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \plus{} 1$ the quotient is $ g(y)$ and the remainder is $ R_2$.  If $ R_1 \equal{} R_2$ then $ m$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \minus{} 1 \qquad \textbf{(E)}\ \text{an undetermined constant}$
1. We start by using polynomial division to find the remainders when \( y^2 + my + 2 \) is divided by \( y - 1 \) and \( y + 1 \).

2. When \( y^2 + my + 2 \) is divided by \( y - 1 \):
   \[
   y^2 + my + 2 = (y - 1)Q_1(y) + R_1
   \]
   To find \( R_1 \), we substitute \( y = 1 \) into the polynomial:
   \[
   1^2 + m(1) + 2 = 1 + m + 2 = m + 3
   \]
   Therefore, \( R_1 = m + 3 \).

3. When \( y^2 + my + 2 \) is divided by \( y + 1 \):
   \[
   y^2 + my + 2 = (y + 1)Q_2(y) + R_2
   \]
   To find \( R_2 \), we substitute \( y = -1 \) into the polynomial:
   \[
   (-1)^2 + m(-1) + 2 = 1 - m + 2 = 3 - m
   \]
   Therefore, \( R_2 = 3 - m \).

4. Given that \( R_1 = R_2 \), we set the remainders equal to each other:
   \[
   m + 3 = 3 - m
   \]

5. Solving for \( m \):
   \[
   m + 3 = 3 - m
   \]
   \[
   m + m = 3 - 3
   \]
   \[
   2m = 0
   \]
   \[
   m = 0
   \]

The final answer is \( \boxed{0} \)