The conditions set the constant and linear <coefficients> to zero, so $\boxed{J_3=(X^2)}$. This is a <principal ideal>, but it is \b[not maximal]: there is a strict chain $(X^2)\subsetneq(X)\subsetneq\mathbb R[X]$. In the <quotient ring>, the nonzero class of $X$ has square zero, so the quotient is not a <field>.

Solution

Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2006 / 3 / iii / Solution

 Solution (source code)