Let $A$ be a (commutative unitary) ring, let $I\subseteq A$ be a non-zero ideal. Are there any injective homomorphisms $A\to A/I$?

I've been thinking for a while, keeping in mind that the homomrphisms $ A/I\to R$ are in natural bijection with the homomorpisms $A\to R$ whose kernels contain $I$ (for any ring $R$), but I couldn't prove anything. On the other hand I tried to find some examples, but without success. Can you give me any suggestion?


1 Answer 1


Sure. The trivial example is letting $I = (0)$ and taking the quotient map. But there are also plenty of examples with $I \neq (0)$.

The first example to come to me would be the following: Let $A = k[x_1, x_2, \dots]$ be the polynomial ring with infinitely many variables and let $I = (x_1)$, so $A/I \cong k[x_2, x_3, \dots]$. The map $$k[x_1, x_2, \dots] \to k[x_2, x_3, \dots], \ x_i \mapsto x_{i+1}$$ is an isomorphism (and in particular injective).


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