# If $f\colon P \hookrightarrow R$ and $(f(a_1), \ldots, f(a_n)) = 1$, then is $(a_1, \ldots, a_n) = 1$?

Let $P$ and $B$ be two unital rings and let $f\colon P\to R$ be a unital injective ring homomorphism. Suppose that $a_1, \ldots, a_n \in P$ are elements such that

$R=R\cdot f(a_1) + \ldots + R\cdot f(a_n)$.

Do we have

$P = P\cdot a_1 + \ldots + P \cdot a_n$ ?

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Fields have the interesting property that every non-zero element generates the unit ideal, and yet there are lots of homomorphisms from things that aren't fields into fields. For concreteness, think about the inclusion of $\mathbf Z$ in $\mathbf Q$. You only need a single $a_1$ to figure out that something is amiss with the statement.