Small provides an example of a ring which is right but not left hereditary is the ring $R =\left( \begin{matrix} \mathbb{Z} & \mathbb{Q} \\ 0 & \mathbb{Q} \end{matrix} \right)$;
To prove that it's not left hereditary he proves that the left ideal $I = \left( \begin{matrix} 0 & \mathbb{Q} \\ 0 & 0 \end{matrix} \right)$ is not $R$-projective.
Here's the proof I found, which I can't quite understand:
If $I$ is projective then reducing modulo the kernel $J$ of the homomorphism $\varphi : R \rightarrow \mathbb{Z}$ defined by $\varphi \left( \left( \begin{matrix} r & p \\ 0 & q \end{matrix} \right) \right) =r $, we get $ I/ J \cdot I \cong (0) \oplus \mathbb{Q} \cong \mathbb{Q}$ which would be projective as a $\mathbb{Z}$-module; which is a contradiction.
Now I know that $\mathbb{Q}$ is not projective over $\mathbb{Z}$;
what I do not get is why $ I/ J \cdot I \cong (0) \oplus \mathbb{Q}$ and
why it should be projective over $\mathbb{Z}$ if $I$ is projective over $R$.
$I/J\cdot I = (0) \oplus \mathbb Q$ is a very strange way to write it, but here's what's going on:
Given any ring $R$, any ideal $J$, and any $R$-module $M$ we have the following:
You can give a fairly elementary proof of this by showing that if $R^n = M \oplus Q$ then $(R/J)^n = M/JM \oplus Q/JQ$ (factor the obvious maps). Alternatively we have a split short exact sequence $$M \to R^n \to Q$$ and additive functors take split SES's to split SES's so $$M \otimes_R R/J \to (R^n) \otimes_R R/J \to Q \otimes_R R/J$$ is split and $M \otimes_R R/J \simeq M/JM$ is a direct summand of $(R^n) \otimes_R R/J \simeq (R/J)^n$. Hence it is projective as an $R/J$-module.
To finish let $M = I$ and $J$ the kernel as stated. Observe that $I\cdot J = (0)$ so if $M = I$ is projective as an $R$-module then $M/JM = I$ would be projective as an $R/J \simeq \mathbb Z$-module.