Definition of a Ring

Rings

Definition 1 A monoid object in the symmetric monoidal category \((\Ab,\otimes, \mathbb{Z})\) is called a ring. (§Abelian Groups, ⁋Theorem 14)

Following the convention when dealing with abelian groups, we write the operation of \(A\) as \(+\). In this situation, for a ring \(A\), the multiplication \(\mu:A\otimes A \rightarrow A\) is denoted by \(\cdot\), and when there is no danger of confusion we may omit it, writing \(\alpha\cdot \beta\) simply as \(\alpha\beta\). Then from

\[\Hom_\Ab(A\otimes A,A)\cong\Bilin(A,A;A)\]

we see that \(\mu\) is bilinear. That is,

\[(\alpha+\beta)\gamma=\alpha\gamma+\beta\gamma,\quad \alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma\]

hold. Also, \(\mu\) satisfies associativity, and

\[\eta:\mathbb{Z}\rightarrow A\]

determines, via the image of \(1\in \mathbb{Z}\), \(A\)’s (multiplicative) identity \(1\). That is, for a set \(A\) to be a ring means that there exist two binary operations \(+,\cdot\) and elements \(0,1\) such that the following conditions hold.

1. \((A, +, 0)\) is an abelian group.
2. \((A,\cdot,1)\) is a monoid.
3. The distributive law holds between \(\cdot\) and \(+\).

Depending on the author, the existence of a multiplicative identity may be omitted from the definition of a ring, but we call such a case a pseudo-ring, or an rng (a ring without \(i\)). In any case, in most cases we will use commutative rings (with unity).

Proposition 2 For a ring \(A\), for any elements \(\alpha,\beta\), the following hold.

1. \(\alpha0=0\alpha=0\),
2. \(\alpha(-\beta)=(-\alpha)\beta=-(\alpha\beta)\).

Proof

1. Since \(0\) is the additive identity, from the equation

   \[0\alpha=(0+0)\alpha=0\alpha+0\alpha\]

   we obtain \(0\alpha=0\). Similarly we can obtain \(\alpha0=0\).

2. By the result of part 1,

   \[0=\alpha0=\alpha(\beta+(-\beta))=\alpha\beta+\alpha(-\beta)\]

   and therefore \(-(\alpha\beta)=\alpha(-\beta)\). Similarly we also obtain \((-\alpha)\beta=-(\alpha\beta)\).

In general, we do not assume \(0\neq 1\) in Definition 1, but if \(0=1\), then by part 1 of Proposition 2, for any \(\alpha\in A\),

\[\alpha=\alpha\cdot 1=\alpha\cdot 0=0\]

would hold, so we must have \(A=\{0\}\). Such a ring is called the zero ring.

Ring Homomorphisms

Definition 3 For two rings \(A,B\), a function \(\phi:A \rightarrow B\) is called a ring homomorphism if for any \(\alpha,\beta\in A\),

\[\phi(\alpha+\beta)=\phi(\alpha)+\phi(\beta),\quad \phi(\alpha\beta)=\phi(\alpha)\phi(\beta),\quad \phi(1)=1\]

hold.

With these as morphisms,

• the category \(\Ring\) of rings,
• the category \(\cRing\) of commutative rings,

and so on are defined. When the existence of an identity of a ring is not assumed, that is, in the case of rngs, the condition \(\phi(1)=1\) in Definition 3 is not required, and the category of rngs with these as morphisms is denoted by \(\Rng\). In \(\Ring\) and \(\cRing\), \(\mathbb{Z}\) is the initial object and \(\{0\}\) is the terminal object, while in \(\Rng\), \(\{0\}\) is the zero object.

By definition, a ring homomorphism \(\phi:A\rightarrow B\) is also a group homomorphism between the two abelian groups \((A,+,0)\) and \((B,+,0)\). Then, for a ring homomorphism \(\phi\), the kernel \(\ker \phi\) is defined as the kernel of this group homomorphism. That is, \(\ker \phi=\phi^{-1}(0)\). Then by §Group Homomorphisms, ⁋Proposition 3, \(\phi\) being injective is equivalent to \(\ker \phi=\{0\}\).

Free Rings over Abelian Groups

By definition, a ring is an abelian group, and a ring homomorphism is also a homomorphism of abelian groups. Thus there exists a forgetful functor \(U:\Ring \rightarrow \Ab\), which is simply the functor that forgets the multiplicative structure. This functor is a right adjoint, and its left adjoint \(F: \Ab \rightarrow \Ring\) is given by the following graded abelian group:

\[F(G)=\bigoplus_{n\geq 0} G^{\otimes n}\]

Here, an element of \(G^{\otimes n}\), \(\alpha_n\), can be written in the form

\[\alpha_n=\sum_{i\in I} \alpha^i_{n1}\otimes\cdots\otimes \alpha^i_{nn},\qquad \text{$\alpha_{nj}^i\in G$, $I$ finite}\]

and the elements of \(F(G)\) are of the form

\[(\alpha_0,\alpha_1,\ldots )=(\alpha_0,0,\ldots)+(0,\alpha_1,\ldots)+\cdots,\qquad \text{$\alpha_n=0$ for all but finitely many $n$}\]

However, since for each \(\alpha_n\), by looking at how many elements’ tensor product the constituent \(\alpha^i_{n1}\otimes\cdots\otimes \alpha^i_{nn}\) are formed of, one can tell which \(G^{\otimes n}\) it belongs to, we may abuse notation and write \((0,\ldots, 0, \alpha_n,0,\ldots)\) as \(\alpha_n\), and then any element of \(F(G)\) can be written in the form

\[\sum_{i\in I} \alpha_{i1}\otimes \cdots\otimes \alpha_{in_i}\]

Now we must define multiplication on \(F(G)\). If multiplication is well-defined, then by the distributive law,

\[\left(\sum_{i\in I} \alpha_{i1}\otimes \cdots\otimes \alpha_{in_i}\right)\left(\sum_{j\in J} \beta_{j1}\otimes \cdots\otimes \beta_{jn_j}\right)=\sum_{(i,j)\in I\times J}(\alpha_{i1}\otimes\cdots \otimes \alpha_{in_i})(\beta_{j1}\otimes\cdots\otimes \beta_{jn_j})\]

must hold. Conversely speaking, if we only define the products of \(\alpha_{i1}\otimes\cdots \otimes \alpha_{in_i}\) and \(\beta_{j1}\otimes\cdots\otimes \beta_{jn_j}\), then the product of all elements of \(F(G)\) is defined through the formula above. And we define

\[(\alpha_{i1}\otimes\cdots \otimes \alpha_{in_i})(\beta_{j1}\otimes\cdots\otimes \beta_{jn_j})=\alpha_{i1}\otimes\cdots\otimes \alpha_{in_i}\otimes \beta_{j1}\otimes\cdots\otimes \beta_{jn_j}\]

By the coherence theorem of [Category Theory] §Monoidal Categories, §§Monoidal Category, one can see that this defines a ring structure on \(F(G)\), and the additive identity of \(F(G)\) is \(0=(0,0,\ldots)\) while the multiplicative identity is \(1=(1,0,\ldots)\). It is not difficult to prove the functoriality of \(G\mapsto F(G)\), and in addition to this, the following holds.

Proposition 4 For \(F\) defined above and the forgetful functor \(U:\Ring \rightarrow \Ab\), we have \(F\dashv U\).

Proof

Let any ring \(A\) and abelian group \(G\) be given. Then we must prove

\[\Hom_\Ring(F(G), A)\cong \Hom_\Ab(G, U(A))\]

For any ring homomorphism \(\phi: F(G) \rightarrow A\), composing with the inclusion \(i:G\hookrightarrow F(G)\) yields an abelian group homomorphism \(\phi\circ i:G \rightarrow U(A)\).

Conversely, if an arbitrary abelian group homomorphism \(f:G \rightarrow U(A)\) is given, the following formula:

\[\sum_{i\in I} \alpha_{i1}\otimes \cdots\otimes \alpha_{in_i}\mapsto \sum_{i\in I} f(\alpha_{i1})\cdots f(\alpha_{in_i})\]

defines a ring homomorphism \(\phi:F(G) \rightarrow A\).

One can verify that the two maps \(\Hom_\Ring(F(G), A) \rightarrow\Hom_\Ab(G, U(A))\) and \(\Hom_\Ab(G, U(A))\rightarrow \Hom_\Ring(F(G), A)\) defined in this way are mutually inverse, and that this bijection is natural.

Subrings and Ideals

Definition 5 For a ring \((A,+,-,\cdot,0,1)\), a subset \(S\) is called a subring if \((S,+,-,\cdot,0,1)\) has a ring structure.

Meanwhile, the following holds.

Proposition 6 For any ring homomorphism \(\phi:A \rightarrow B\), \(\ker \phi\) is a subgroup of \((A,+,0)\) and is closed under multiplication.

Proof

Since we have verified that \(\ker \phi\) is a subgroup of the abelian group \((A,+,0)\), it suffices to show that \(\ker \phi\) is closed under multiplication. But for any \(\alpha,\beta\in\ker \phi\),

\[\phi(\alpha\beta)=\phi(\alpha)\phi(\beta)=0\cdot 0=0\]

so \(\alpha\beta\in\ker \phi\) holds.

If we examine the above proof closely, we can verify that even if only one of the two elements \(\alpha,\beta\) belongs to \(\ker \phi\), \(\alpha\beta\) belongs to \(\ker \phi\). We define this as follows.

Definition 7 Let a ring \(A\) be given. Then \(\mathfrak{a}\subseteq A\) is called a left ideal (resp. right ideal) if \(\mathfrak{a}\) is a subgroup of \((A,+,0)\) and for any \(x\in\mathfrak{a}\) and \(\alpha\in A\), we have \(\alpha x\in\mathfrak{a}\) (resp. \(x\alpha\in\mathfrak{a}\)).

If \(\mathfrak{a}\) is both a left ideal and a right ideal, it is called a two-sided ideal.

In particular, the above observation says that for any ring homomorphism \(\phi\), \(\ker\phi\) is a two-sided ideal of \(A\).

For convenience, from now on we will only prove propositions about left ideals (or propositions about two-sided ideals), but all propositions about left ideals also hold for right ideals with appropriate modifications. In any case, since most of the rings we actually use are commutative, the distinction between left ideals, right ideals, and two-sided ideals is not necessary.

It is easy to prove that the intersection of left ideals is a left ideal. Meanwhile, in \(A\), for any element \(x\), the following set:

\[Ax=\{\alpha x\mid\alpha\in A\}\]

is a subgroup under addition, and furthermore, for any \(\beta\in A\) and \(\alpha x\in Ax\), since \(\beta(\alpha x)=(\beta\alpha)x\in Ax\), \(Ax\) is a left ideal of \(A\). One can check that this in fact coincides with the smallest left ideal containing \(x\), that is, the intersection of all left ideals containing \(x\), and the same argument applies to right ideals and two-sided ideals.

More generally, in \(A\), if we define the sum \(\mathfrak{a}+\mathfrak{b}\) of left ideals as the following set:

\[\mathfrak{a}+\mathfrak{b}=\{x+y\mid x\in \mathfrak{a},y\in \mathfrak{b}\}\]

then \(\mathfrak{a}+\mathfrak{b}\) is again a left ideal, and in fact this is the smallest left ideal containing the two left ideals \(\mathfrak{a}\) and \(\mathfrak{b}\). Then, in \(A\), for any elements \(x_1,\ldots, x_n\), the following ideal:

\[Ax_1+\cdots+Ax_n\]

is the smallest of all left ideals containing \(x_1,\ldots, x_n\). Similarly, one can also define

\[x_1A+\cdots+x_nA,\qquad Ax_1A+\cdots +Ax_nA\]

where \(AxA\) is the finite sums of elements of the form \(\alpha x\beta\). Then the above two sets are the smallest right ideal and two-sided ideal containing \(x_1,\ldots, x_n\), respectively. If \(A\) is commutative, then the notions of left ideal, right ideal, and two-sided ideal all coincide, so we sometimes lump them together and write \((x_1,\ldots, x_n)\).

Meanwhile, for any left ideal \(\mathfrak{a}\), the condition \(1\in\mathfrak{a}\) is equivalent to \(\mathfrak{a}=A\). Therefore, for \(\mathfrak{a}\subsetneq A\), we must have \(1\not\in\mathfrak{a}\).

Subrings and ideals are generally not the same concept. First, since a subring always contains \(1\) by definition, if some subring is simultaneously an ideal then, as just observed, it must be all of \(A\). Even if we weaken the notion of subring to \(\Rng\) and do not require the inclusion of \(1\), the condition that an ideal must be closed under multiplication by arbitrary elements is not something that a subring satisfies, so the two concepts remain distinct. Thus it is a natural question to ask for an example of a subring that is not an ideal, and the most important example is the center of a ring.

Definition 8 For a ring \(A\), the center \(Z(A)\) is the set of elements that commute with all elements of \(A\), that is,

\[Z(A)=\{z\in A\mid za=az\text{ for all }a\in A\}\]

\(Z(A)\) is closed under addition and multiplication and contains \(1\), so it is a subring of \(A\), and is commutative by definition. However, \(Z(A)\) is generally not an ideal. For \(x\in Z(A)\) and \(a\in A\), let us see whether \(ax\) still belongs to the center. For any \(y\in A\),

\[(ax)y=a(xy)=a(yx)\]

and since the element \(x\) of the center commutes with \(y\) but \(a\) does not, \(ax\) need not be an element of the center. In fact, if we set \(x=1\), then \(ax=a\), and unless \(a\) is central, for some \(y\) we have \(ay\neq ya\), that is, \((ax)y\neq y(ax)\), so that \(ax=a\not\in Z(A)\). Therefore \(Z(A)\) does not satisfy the condition for an ideal, and \(Z(A)\) is an ideal only when \(Z(A)=A\), that is, when \(A\) is commutative. Thus, in a noncommutative ring, \(Z(A)\) is a typical example of a subring that is not an ideal.

Definition 9 For a ring \(A\) and an ideal \(\mathfrak{m}\subsetneq A\), if there does not exist, satisfying \(\mathfrak{m}\subsetneq\mathfrak{a}\subsetneq A\), an ideal \(\mathfrak{a}\), then \(\mathfrak{m}\) is called a maximal ideal.

For \(A\), suppose any ideal \(\mathfrak{a}\subsetneq A\) is given. Then we can consider the collection of ideals containing \(\mathfrak{a}\) and distinct from \(A\) itself among ideals of \(A\). This collection is an inductive set, so it has a maximal element. ([Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma)) It is not difficult to see that this maximal element is a maximal ideal, and thus we obtain the following.

Theorem 10 (Krull) For a ring \(A\) and an ideal \(\mathfrak{a}\subsetneq A\), there always exists, containing \(\mathfrak{a}\), a maximal ideal of \(A\), \(\mathfrak{m}\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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