Invariant dimension property and a ring epimorphism

Question

In Hungerford's Algebra, p. 186, the Proposition 2.11 says

Let $f:R\to S$ be a nonzero epimorphism of rings with identity. If $S$ has the invariant dimension property (IDP), then so does $R$.

It is true, and the proof is logical. However, my problem is about the converse;

if $R$ has the IDP, then does $S$ have the IDP too?

We do not need to handle infinite basis cases or commutative ring cases, because that is true... (Theorem 2.6 Corollary 2.12).

Answer 1

Let $S$ be any ring without IBN, and let $T$ be any ring with IBN.

There is a projection $S\times T \to T$, so $S\times T$ has $IBN$. But then the projection $S\times T \to S$ is a counterexample.