Editing Kernel (algebra) (section)

=== Ring homomorphisms ===
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Let ''R'' and ''S'' be [[ring (mathematics)|ring]]s (assumed [[unital algebra|unital]]) and let ''f'' be a [[ring homomorphism]] from ''R'' to ''S''.
If 0<sub>''S''</sub> is the [[zero element]] of ''S'', then the ''kernel'' of ''f'' is its kernel as additive groups.<ref name=":1" /> It is the preimage of the [[zero ideal]] {{mset|0<sub>''S''</sub>}}, which is, the subset of ''R'' consisting of all those elements of ''R'' that are mapped by ''f'' to the element 0<sub>''S''</sub>.
The kernel is usually denoted {{nowrap|ker ''f''}} (or a variation).
In symbols:
: <math> \operatorname{ker} f = \{r \in R : f(r) = 0_{S}\} .</math>

Since a ring homomorphism preserves zero elements, the zero element 0<sub>''R''</sub> of ''R'' must belong to the kernel.
The homomorphism ''f'' is injective if and only if its kernel is only the singleton set {{mset|0<sub>''R''</sub>}}.
This is always the case if ''R'' is a [[field (mathematics)|field]], and ''S'' is not the [[zero ring]].

Since ker ''f'' contains the multiplicative identity only when ''S'' is the zero ring, it turns out that the kernel is generally not a [[subring]] of ''R.'' The kernel is a sub[[rng (algebra)|rng]], and, more precisely, a two-sided [[ideal (ring theory)|ideal]] of ''R''.
Thus, it makes sense to speak of the [[quotient ring]] {{nowrap|''R'' / (ker ''f'')}}.
The first isomorphism theorem for rings states that this quotient ring is naturally isomorphic to the image of ''f'' (which is a subring of ''S''). (Note that rings need not be unital for the kernel definition).