Ring Homomorphisms and Quotients

Let R be a ring, $S \subseteq R$ is a subring of R $(S \leq R$) if 1) $(S, +) \leq (R, +)$ (equivalently, S is closed under subraction and nonempty) 2) for all $x, y \in S xy \in S$

Remark: if R is unitary we also want $1_R \in S$, in which case we say S is a unitary subring.

Examples: $\mathbb{Z} \leq \mathbb{Q} \leq \mathbb{R} \leq \mathbb{C}$ (unitary subrings). $n\mathbb{Z} \leq \mathbb{Z}$ is a subring but is not unitary unless $n = \pm 1$.

Let R and S be rings and $\phi : R \rightarrow S$ is a function. If for all $x, y \in R \phi(x+y) = \phi(x) + \phi(y)$ (it is a ring homomorphism) and for all $x, y \in R \phi(xy) = \phi(x)\phi(y)$ then $\phi$ is a ring homomorphism.

Remarks: if R and S are unitary we might want $\phi(1_R) = 1_S$ (in that case we will talk of a homomorphism of unitary rings). A bijective ring homomorphism is called an isomorphism.

Proposition: Let R and S be rings and $\phi: R \rightarrow S$ be a ring homomorphism. 1) for all $U \leq R, \phi(u) \leq S$ 2) for all $V \leq S, \phi^{-1}(V) \leq R$

Let $I \subseteq R$ (where R is a ring). I is an ideal if 1) $(I, +) \leq (R, +)$ are groups 2) for all $x \in I, y \in R xy \in I, yx \in I$ When R is unitary and $I \leq R$ is an ideal, if $1 \in I$ then I = R (in fact if $I \cap R*$ is nonempty then I = R)

Proposition: let R and S be rings and $\phi : R \rightarrow S$ a ring hommorphism, then ker($\phi$) is an ideal in R.

Proof: $(ker(\phi), +) \leq (R,+)$ by definition. For all $x \in ker(\phi), y \in R \phi(xy) = \phi(x)\phi(y) = 0 \cdot \phi(y) = 0$ hence $xy \in ker(\phi)$. $\phi(xy) = \phi(y)\phi(x) = \phi(y)\cdot 0$ hence $yx \in ker(\phi)$.

Proposition: let R be a ring and $I \subseteq R$ an ideal. The group $(R/I, +)$ can be made into a ring such that the map $\pi : R \rightarrow R/I$ that sends $x \mapsto x+I$ is a ring homomorphism.

Proof: $(x+I) \cdot (y+I) = xy + I$ (where $x,y \in R$). Well defined? let $x, y, z \in R$ and $s, t \in I$, then $(x+s)(y+t) = xy + sy + xt + st \in I$ (by definition of ideals). Associativity: $((x+I)(y+I))(z+I) = (xy+I)(z+I) = xyz+I = (x+I)((y+I)(z+I))$ . Distributivity: $(x+I)((y+I)+(z+I)) =$ $(x+I)(y+z+I) =$ $x(y+z)+I =$ $xy+xz+i = (x+I)(y+I)+(x+I)(z+I)$ (the proof of right distributivity is left as an exercise). So $(R/I, +, \cdot)$ is a ring (we call R/I the quotient ring of R modulo I).

Note that if R is unitary and $I \subseteq R$ is an ideal then R/I is unitary (similarly if R is commutative, R/I is commutative).

Propositon: let R be a ring, $I \subseteq R$. The following are equivalent: 1) I is an ideal 2) there exists a $\phi: R \rightarrow S$ such that $I = ker(\phi)$

Proof: for $1 \rightarrow 2: \phi = \pi: R \rightarrow R/I, ker(\phi) = I$. $2 \rightarrow 1$ was proved above ($ker(\phi)$ is an ideal in R)

First isomorphism theorem (for rings): let R and S be rings and $\phi : R \rightarrow S$ is a ring homomorphism. Then $\phi(R) \cong R/ker(\phi)$ (the proof is left as an exercise to the reader, as it is very similar to the proof of this theorem for groups).