Intro to Ring Theory

Refer Dummit & Foote (2004) chapter 7
Additional material from Aluffi (2009) chapter 3
These references disagree on many definitions.

Preliminaries

Definition 68.1(Ring).

A ring $R$ is a set with two binary operations, $+$ and $\cdot$, satisfying

1. $(R,+)$ is an abelian group.
2. $\cdot$ is associative: $(ab)c=a(bc)$ for all $a,b,c\in R$.
3. the distributive law holds in $R$: for all $a,b,c\in R$, $(a+b)c=ac+bc$ and $a(b+c)=ab+ac$.

$R$ is said to be commutative if $\cdot$ is commutative. $R$ is said to have an identity if there is an element $1\in R$ with $1a=a1=a$ for all $a\in R$.

Warning

Many sources (including Conrad’s notes) require that a ring $R$ has an identity (in other words, that $(R,\cdot )$ be a monoid).

Note that if $R$ has an identity, the distributive law forces $(R,+)$ to be abelian:

$$\begin{array}{rl}(1+1)(a+b)& =(1+1)a+(1+1)b\\ & =a+a+b+b\\ & \\ (1+1)(a+b)& =1(a+b)+1(a+b)\\ & =a+b+a+b\\ & \\ ⟹& a+b=b+a.\end{array}$$

Proposition 68.2(Properties of rings).

Let $R$ be a ring. Then,

1. $0a=a0=0$ for all $a\in R$.
2. $(-a)b=a(-b)=-ab$ for all $a,b\in R$.
3. $(-a)(-b)=ab$ for all $a,b\in R$.
4. If $R$ has an identity, it is unique.
5. If $R$ has an identity and $a\in R$ has a multiplicative inverse (that is, there exists $b\in R$ such that $ab=ba=1$), it is unique.

Example 68.3(The zero and trivial rings).

If $R$ has an identity and $1=0$, $a=1a=0a=0$ for all $a\in R$. Thus, $R=\{0\}$, the zero ring. We usually exclude this possibility by requiring that $1\ne 0$.

Let $(R,+)$ be an abelian group. Define $ab=0$ for all $a,b\in R$. This is a ring, and is called the trivial ring. Multiplication does not add any new structure, and the trivial ring does not have an identity unless it is the zero ring.

Definition 68.4(Definition).

1. A nonzero element $a\in R$ is a left-zero-divisor if there exist elements $b\ne 0$ in $R$ for which $ab=0$. Ditto for right zero divisors. Both are collectively called zero divisors.
2. Assume $R$ has an identity $1\ne 0$. An element $u\in R$ is a left-unit if $\exists v\in R$ such that $uv=1$.; it is a right-unit if $\exists v\in R$ such that $vu=1$. Units are two-sided units.

Proposition 68.5(Aluffi III.1.12).

In a ring $R$,

1. $u$ is not a left (right) zero divisor $⟺$ left (right) multiplication by $u$ is injective;
2. $u$ is a left (right) unit $⟺$ left (right) multiplication by $u$ is surjective;
3. $u$ is a left (right) unit $⟹$ right (left) multiplication by $u$ is injective, that is, $u$ is not a right (left) zero divisor.

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It follows that a two sided zero divisor can not be a two sided unit.

Definition 68.6(Definition).

1. A ring $R$ with identity $1\ne 0$ is called a division ring if every nonzero $a\in R$ is a unit.
2. A commutative ring $R$ with identity $1\ne 0$ is called an integral domain if it has no zero divisors.

A commutative division ring is a field. A finite integral domain is also a field (Immediate proof from Proposition 5). Another equivalent way to arrive at a field is to make $(R^{\times },\cdot )$ an abelian group for any ring $R$.

Theorem 68.7(Wedderburn).

A finite division ring is necessarily commutative, i.e, is a field.

Definition 68.8(Definition).

A subring of the ring $R$ is a subgroup of $R$ that is closed under multiplication.

Note

Let $S\subseteq R$ be a subring. Let $1_S$ and $1_R$ be the identities of $S$ and $R$ respectively. Then,

$$\begin{array}{r}{1}_S{1}_R=1_S\\ 1_S{1}_S=1_S\\ ⟹1_S(1_S-1_R)=0\end{array}$$

If $R$ is an integral domain, $1_S=1_R$ is forced. Also, if we require in our definition of ring homomorphisms for identity to be mapped to identity, the inclusion map $S\to R$ forces $1_S=1_R$. In this course, you can assume $1_S=1_R$.

To show that a subset of a ring $R$ is a subring, it suffices to show that it is nonempty and closed under subtraction and multiplication.

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Ring homomorphisms and quotient rings

Definition 68.9(Definition).

Let $R$ and $S$ be rings.

1. A ring homomorphism is a map $\phi :R\to S$ satisfying $\phi (a+b)=\phi (a)+\phi (b)$ for all $a,b\in R$ and $\phi (ab)=\phi (a)\phi (b)$ for all $a,b\in R$.
2. The kernel of the ring homomorphism $\phi$, denoted $\ker \phi$, is the set of elements of $R$ that map to $0$ in $S$.
3. A bijective ring homomorphism is called an isomorphism.

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If $\phi :R\to S$ is a ring homomorphism, $\phi :(R,+)\to (S,+)$ is a group homomorphism, and their kernels coincide.

Proposition 68.10(Proposition).

Let $R$ and $S$ be rings and let $\phi :R\to S$ be a homomorphism.

1. $\mathrm{Im}\phi$ is a subring of $S$.
2. $\ker \phi$ is a subring of $R$. Furthermore, if $\alpha \in \ker \phi$, then $r\alpha$ and $\alpha r\in \ker \phi$ for every $r\in R$, i.e, $\ker \phi$ is closed under multiplication by elements from $R$.

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Analogous to this result from group theory.

Ideals

Let $\phi :R\to S$ be a ring homomorphism with kernel $I$. Since $\phi$ is in particular a homomorphism of abelian groups, the fibers of $\phi$ (the additive cosets $r+I$, denoted by $R/I$) have the additive group structure of the additive quotient group of the additive abelian group $R$ by the (necessarily normal) subgroup $I$, with addition defined by

$$(r+I)+(s+I)=(r+s)+I.$$

Additionally, we can define a multiplicative structure on $R/I$ by

$$(r+I)\times (s+I)=(rs)+I.$$

This is well defined: for $\alpha ,\beta \in I$, we have (from Proposition 10)

$$\begin{array}{rl}(r+\alpha +I)\times (s+\beta +I)& =(r+\alpha )(s+\beta )+I\\ & =rs+I.\end{array}$$

This makes the additive quotient group $R/I$ into a ring. Distributivity follows directly from the distributivity of $R$.

As in the case for groups, we can also consider when $+$ and $\times$ as defined above can be used to define a ring structure on the collection of cosets of an arbitrary subgroup $I$ of $R$ (for groups, we found that for $H<G$, $G/H$ is a group with the group operation defined by $(aH)(bH)=(ab)H$ iff $H◃G$). As noted above, since $R$ is an abelian additive group, the subgroup $I$ is necessarily normal, so $R/I$ is automatically an additive abelian group. However, $R/I$ in general does not have a multiplicative structure induced from $R$. So, what does it take?

Let $I$ be an arbitrary subgroup of $(R,+)$. For multiplication as defined to be well defined, we must have $(r+\alpha )(s+\beta )+I=rs+I$ for all $r,s\in R$ and all $\alpha ,\beta \in I$. Letting $r=s=0$, we see that $I$ must be closed under multiplication, that is, $I$ must be a subring of $R$. Next, by letting $s=0$, we see that $I$ must be closed under multiplication on the left by elements from $R$. Similarly, letting $r=0$ tells us that $I$ must be closed under multiplication on the right by elements from $R$. Note that these are precisely the properties we used from Proposition 10 above to show that $R/I$ is a ring when $I$ is the kernel of a ring homomorphism. Thus, the quotient $R/I$ of a ring $R$ by a subgroup $I$ has a natural ring structure iff $I$ is closed under multiplication on the left and right by the elements from $R$ (so in particular must be a subring of $R$). Such subrings are called ideals.

Definition 68.11(Ideal).

Let $R$ be a ring, and let $I$ be a subgroup of $R$.

1. $I$ is a left ideal of $R$ if $I$ is a subring of $R$ and $I$ is closed under left multiplication by elements from $R$, i.e, $rI\subseteq I$ for all $r\in R$.
2. $I$ is a right ideal of $R$ if $I$ is a subring of $R$ and $I$ is closed under right multiplication by elements of $R$.
3. $I$ is an ideal of $R$ if it is both a left ideal and a right ideal of $R$.

Proposition 68.12(Proposition).

Let $R$ be a ring and let $I$ be an ideal of $R$. Then the additive quotient group $R/I$ is a ring under the binary operations

$$\begin{array}{rl}& (r+I)+(s+I)=(r+s)+I\\ & (r+I)\times (s+I)=(rs)+I\end{array}$$

for all $r,s\in R$. Conversely, if $I$ is any subgroup such that the above operations are well defined, then $I$ is an ideal of $R$.

The isomorphism theorems

Theorem 68.13(First isomorphism theorem for rings).

If $\phi :R\to S$ is a ring homomorphism, then $\ker \phi$ is an ideal of $R$, $\phi (R)$ is a subring of $S$, and $R/\ker \phi \cong \phi (R)$. *

Theorem 68.14(Theorem).

If $I$ is an ideal of $R$, them the map $R\to R/I$ defined by $r\mapsto r+I$ is a surjective ring homomorphism with kernel $I$, called the natural projection.

Thus, every ideal is the kernel of a ring homomorphism and vice versa *.

Theorem 68.15(Second isomorphism theorem for rings).

Let $A$ be a subring and let $B$ be an ideal of $R$. Then $A+B=\{a+b|a\in A,b\in B\}$ is a subring of $R$, $A\cap B$ is an ideal of $A$ and $(A+B)/B\cong A/(A\cap B)$. *

Proof.

From the second isomorphism theorem for groups (note that$N_{(R,+)}(B)=(R,+)$), $A+B$ is a subgroup of $R$, $A\cap B$ is a subgroup of $A$, and $(A+B)/B$ and $A/(A\cap B)$ are isomorphic as groups. It is easy to check that $A+B$ is closed under multiplication. Thus, $A+B$ is a subring of $R$. Define a map $A\to (A+B)/B$ by $a\mapsto a+B$. This is a ring homomorphism. It follows from the first isomorphism theorem that $A\cap B$ is an ideal of $A$ and $(A+B)/B\cong A/(A\cap B)$.□

Theorem 68.16(Third isomorphism theorem for rings).

Let $I$ and $J$ be ideals of $R$ with $I\subseteq J$. Then $J/I$ is an ideal of $R/I$ and $(R/I)/(J/I)\cong R/J$. *

Proof.

Note that since$I$ is an ideal of $J$, $J/I$ is a ring. It is easily verified that $J/I$ is closed under multiplication with elements from $R/I$. Thus, $J/I$ is an ideal of $R/I$, and $(R/I)/(J/I)$ is a ring. Consider the map $R/I\to R/J$ defined by $a+I\mapsto a+J$. This is a ring homomorphism with kernel $J/I$. It follows from the first isomorphism theorem that $(R/I)/(J/I)\cong R/J$.□

Theorem 68.17(Correspondence theorem for rings).

Let $I$ be an ideal of $R$. The correspondence $A\leftrightarrow A/I$ is an inclusion preserving bijection between the set of subrings $A$ of $R$ that contain $I$ and the set of subrings of $R/I$. Also, $A$ (a subring containing $I$) is an ideal of $R$ iff $A/I$ is an ideal of $R/I$. *

Proof.

From the correspondence theorem for groups, we have a inclusion preserving bijection between the set of subgroups of$R$ containing $I$ and the set of subgroups of $R/I$, given by $A\leftrightarrow A/I$. Since $A$ is a subring of $R$ $⟺$ $A/I$ is a subring of $R/I$, a restriction of this bijection to the set of subrings of $R$ is the correspondence we are after. The second assertion is trivial.□

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

Dummit, D. S., & Foote, R. M. (2004). Abstract Algebra (3rd ed). Wiley.