Ideal $I$ in a Noetherian ring such that $I+(b)$ and $I+(a)$ are projective, where $ab\in I$ Let $R$ be a Noetherian ring and $I$ be a proper ideal of $R$ and let $a,b\in R$ be such that $ab\in I$ and $I+Ra$ and $I+Rb$ are projective as $R$-modules. 
Then is the ideal $I$ projective as an $R$-module ?
NOTE: If this holds, then it would imply that if in a Noetherian ring, every prime ideal is projective, then every ideal is projective 
 A: This is not true. Let $R = k[x,y,z]/(xy+xz+yz)$ (for $k$ a field). Let $I = \langle xy,xz,yz \rangle$. Geometrically, $R$ is the coordinate ring of a cone in three dimensional space containing the three coordinate axes, and $I$ is the ideal of the three axes. 
We first check that $I$ is not projective. Since $I \otimes \mathrm{Frac}(R)$ is one dimensional as a $\mathrm{Frac}(R)$ vector space, it would have to be of rank $1$. But $I \otimes R/\langle x,y,z \rangle$ is a $2$-dimensional $k$ vector space (it is spanned by $xy$, $xz$ and $yz$, with the relation $xy+xz+yz=0$), so $I$ is not locally free of rank $1$ at $\langle x,y,z \rangle$.
We have $xy$, $xz$ and $yz \in \langle x \rangle$; for the last, note that $yz = -x(y+z)$. So $I + Rx = \langle x \rangle$ which is principal and hence projective. For the same reason, $I+Ry$ is projective. And $xy \in I$.
If you have some algebraic geometry, here is how I thought about this. Let $X = \mathrm{Spec}(R)$ and $D \subset X$ be the zero locus of $I$. Let $D_a$ and $D_b$ be the zero locuses of $I+Ra$ and $I+Rb$. So $D \supseteq D_a$, $D_b$. The equation $ab \in I$ means that $ab$ vanishes on $D$, so $D = D_a \cup D_b$. "Projective" means "locally principal" means "Cartier divisor". So I want $D_a$ and $D_b$ to be Cartier divisors, but $D_a \cup D_b$ to be not Cartier. Since $D_a$ and $D_b$ are Cartier, their union will be pure codimension $1$, in other words, a Weil divisor. So I want a non-Cartier Weil divisor to be the union of two Cartier divisors.
Time to try the simplest ring which has non-Cartier Weil divisors: On the quadratic cone, any number of lines through $0$ is a Weil divisor, but it is only Cartier if the number of lines is even. So I want an odd number of lines (3 in the example) to be the union of two sets with an even number of lines (2 each in the example).
