First isomorphism theorem for rings

The isomorphism is the map \(\bar\varphi(r+\ker\varphi)=\varphi(r)\). It is well defined and injective because \(\varphi(r)=\varphi(s)\) is equivalent to \(r-s\in\ker\varphi\), that is, to \(r\) and \(s\) lying in the same coset; it is onto the image by construction, and it preserves sums and products because \(\varphi\) does.

The kernel of a ring homomorphism is always an ideal of \(R\) (not a subring, since it does not contain 1 unless \(\varphi\) is zero), which is why forming \(R/\ker\varphi\) makes sense. Conversely, every ideal \(I\) is the kernel of the projection \(R\to R/I\); so ideals and kernels of homomorphisms are exactly the same thing.

The most common case is \(\mathbb{Z}\to\mathbb{Z}_n\), which gives \(\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n\). When \(\varphi\) is surjective the image is \(S\) and the theorem reads \(R/\ker\varphi\cong S\), so \(|R|=|\ker\varphi|\,|S|\) if \(R\) is finite. The map \(x\mapsto(x\bmod 2,\,x\bmod 3)\) from \(\mathbb{Z}_6\) to \(\mathbb{Z}_2\times\mathbb{Z}_3\) is an example: its kernel is zero, so it is an isomorphism.

A typical mistake is to claim that \(R/\ker\varphi\) is isomorphic to \(S\) without checking surjectivity. This theorem is the analogue of the first one for groups and is used to identify quotients: in the next section, by evaluating polynomials, it will let us see \(R[x]/(x-a)\cong R\).