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  text: '[summary: The kernel of a [-ring_homomorphism] is the collection of elements which the homomorphism sends to $0$.]\n\nGiven a [-ring_homomorphism] $f: R \\to S$ between [3gq rings] $R$ and $S$, we say the **kernel** of $f$ is the collection of elements of $R$ which $f$ sends to the zero element of $S$.\n\nFormally, it is $$\\{ r \\in R \\mid f(r) = 0_S \\}$$\nwhere $0_S$ is the zero element of $S$.\n\n# Examples\n\n- Given the "identity" (or "do nothing") ring homomorphism $\\mathrm{id}: \\mathbb{Z} \\to \\mathbb{Z}$, which sends $n$ to $n$, the kernel is just $\\{ 0 \\}$.\n- Given the ring homomorphism $\\mathbb{Z} \\to \\mathbb{Z}$ taking $n \\mapsto n \\pmod{2}$ (using the usual shorthand for [-5ns]), the kernel is the set of even numbers.\n\n# Properties\n\nKernels of ring homomorphisms are very important because they are precisely [ideal_ring_theory ideals]. ([5r9 Proof.])\nIn a way, "ideal" is to "ring" as "[-576]" is to "[3gd group]", and certainly [subring_ring_theory subrings] are much less interesting than ideals; a lot of ring theory is about the study of ideals.\n\nThe kernel of a ring homomorphism always contains $0$, because a ring homomorphism always sends $0$ to $0$.\nThis is because it may be viewed as a [-47t] acting on the underlying additive group of the ring in question, and [49z the image of the identity is the identity] in a group.\n\nIf the kernel of a ring homomorphism contains $1$, then the ring homomorphism sends everything to $0$.\nIndeed, if $f(1) = 0$, then $f(r) = f(r \\times 1) = f(r) \\times f(1) = f(r) \\times 0 = 0$.',
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