Rings add multiplication to abelian groups. Ideals are the correct kernels for ring homomorphisms, and quotient rings are the basic way to impose algebraic relations.
Rings and ring homomorphisms
Definition. A ring \(R\) is an abelian group under addition with an associative multiplication distributing over addition. In these notes rings are assumed unital unless stated otherwise. A ring homomorphism \(f:R\to S\) preserves addition, multiplication, and \(1\).
Example. The rings \(\mathbb Z\), \(F[x]\), \(M_n(F)\), and \(\mathbb Z/n\mathbb Z\) are basic. The matrix ring is usually noncommutative, while \(F[x]\) is commutative.
Ideals as kernels
Definition. A two-sided ideal \(I\trianglelefteq R\) is an additive subgroup such that \(rI\subset I\) and \(Ir\subset I\) for all \(r\in R\). For commutative rings, one simply says ideal.
Proposition. The kernel of a ring homomorphism is an ideal, and the image is a subring. Conversely, every ideal \(I\trianglelefteq R\) is the kernel of the quotient map \(R\to R/I\).
Proof. If \(a,b\in\ker f\), then \(f(a-b)=0\). If \(r\in R\), then \(f(ra)=f(r)f(a)=0\) and similarly \(f(ar)=0\). The quotient map sends \(r\) to \(r+I\), whose zero coset is exactly \(I\).
Quotient rings
Theorem. If \(I\trianglelefteq R\), then the additive quotient \(R/I\) is a ring under \[ (r+I)(s+I)=rs+I. \] The projection \(\pi:R\to R/I\) is a surjective ring homomorphism with kernel \(I\).
Proof. If \(r-r'\in I\) and \(s-s'\in I\), then \(rs-r's'=r(s-s')+(r-r')s'\in I\), using ideal absorption. Hence multiplication is well-defined; the ring axioms descend from \(R\).
Polynomial quotient rings
Example. In \(F[x]/(x^2+1)\), the class \(\alpha=x+(x^2+1)\) satisfies \(\alpha^2=-1\). If \(F=\mathbb R\), this quotient is isomorphic to \(\mathbb C\). If \(F=\mathbb C\), then \(x^2+1\) factors, and the quotient is not a field.
Calculation. Every class in \(F[x]/(f)\) has a representative of degree less than \(\deg f\), by division with remainder. Thus \(F[x]/(f)\) is finite-dimensional over \(F\) with basis \(1,x,\ldots,x^{d-1}\) when \(f\) is monic of degree \(d\).
The universal property of quotients
Theorem. If \(f:R\to S\) is a ring homomorphism and \(I\subset\ker f\), then there is a unique homomorphism \(\overline f:R/I\to S\) such that \(f=\overline f\circ\pi\). It is given by \(\overline f(r+I)=f(r)\).
Remark. Quotient rings are not merely sets of congruence classes. They are the universal way of forcing all elements of \(I\) to become zero.