A uniform module as an intersection

Question

Let $R$ be a semiprime, nonsingular ring with finite Goldie dimension u.dim $R_R$. (Nonsingularity means here that $Z(R_R)=0$, where $Z(R_R)$ is the set of elements $x$ of $R$ with $ann(x)$ is essential in $R_R$ (as right $R$-modules)). Let $I$ be a right ideal of $R_R$ with u.dim $I_R$=$n$ . So, by definition of Goldie dimension, we can choose a right ideal $I_0$ of $R$ with $I_0⊆I$ with Goldie dimension $n-1$ (as right $R$-module), and suppose we could pick $a_0∈I_0$ with $ann(a_0)∩I_0=0$.

My question is: why the Goldie dimension of $U_R:=ann(a_0)∩I$ is equal to $1$ if it is nonzero? Thanks in advance for answering.

Answer 1

Another useful characterization of uniform dimension is this:

$u.dim(M)$ is the supremum of the set $\{k\mid M \text{ contains a direct sum of $k$ nonzero submodules}\}$

(This can be found on page 214 in Lam's Lectures on modules and rings.)

This being the case, take a direct sum of $n-1$ submodules of $I_0$ and call it $U_1$. If $U_R:=ann(a_0)\cap I$ contained a direct sum of two or more nonzero submodules, then you could tack those two submodules onto $U_1$ into a bigger direct sum (since $ann(a_0)\cap I_0=\{0\}$) to get a direct sum of strictly more than $n$ nonzero submodules of $I$. This contradicts the fact $I$ has right uniform dimension $n$.