Definition of a Ring

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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 for abelian groups, we write the operation of \(A\) as \(+\). In this setting, the multiplication \(\mu:A\otimes A \rightarrow A\) of a ring \(A\) is denoted by \(\cdot\), and when there is no danger of confusion we 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. Moreover, \(\mu\) is associative, and

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

determines the (multiplicative) identity \(1\) of \(A\) via the image of \(1\in \mathbb{Z}\). In other words, 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 \(+\).

Some authors omit the existence of a multiplicative identity from the definition of a ring, but we call such an object a pseudo-ring, or an rng (a ring without \(i\)). In any case, we will mostly work with commutative rings (with unity).

Proposition 2 For any elements \(\alpha,\beta\) of a ring \(A\), 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 identity

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

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

2. By 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.

Taking these as morphisms, we define

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

and so on. 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 \((B,+,0)\). The kernel \(\ker \phi\) of a ring homomorphism \(\phi\) is defined as the kernel of this group homomorphism. That is, \(\ker \phi=\phi^{-1}(0)\). Then by §Group Homomorphisms, ⁋Proposition 3, \(\phi\) is injective if and only if \(\ker \phi=\{0\}\).

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 simply 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 \(\alpha_n\) of \(G^{\otimes n}\) 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 each \(\alpha_n\) can be identified as belonging to some \(G^{\otimes n}\) by counting how many tensor factors appear in the constituents \(\alpha^i_{n1}\otimes\cdots\otimes \alpha^i_{nn}\), we may abuse notation and write \((0,\ldots, 0, \alpha_n,0,\ldots)\) as \(\alpha_n\); 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, if we define only 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 by the above formula. We set

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

By the coherence theorem of [Category Theory] §Monoidal Categories, §§Monoidal Category, 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 moreover the following holds.

Proposition 4 For the \(F\) 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, given any abelian group homomorphism \(f:G \rightarrow U(A)\), the 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 A subset \(S\) of a ring \((A,+,-,\cdot,0,1)\) is called a subring if \((S,+,-,\cdot,0,1)\) carries a ring structure.

On the other hand, 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 already 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 see that even if only one of the two elements \(\alpha,\beta\) belongs to \(\ker \phi\), the product \(\alpha\beta\) still lies in \(\ker \phi\). We encode this in the following definition.

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 simultaneously a left ideal and a right ideal, we call it a two-sided ideal.

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

For convenience, we will only prove propositions about left ideals (or two-sided ideals) from now on, but all propositions about left ideals also hold for right ideals with appropriate modifications. In any case, most rings we will actually use are commutative, so the distinction between left ideals, right ideals, and two-sided ideals is unnecessary.

It is easy to prove that the intersection of left ideals is a left ideal. On the other hand, for any element \(x\) of \(A\), the set

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

is a subgroup under addition, and moreover for any \(\beta\in A\) and \(\alpha x\in Ax\), we have \(\beta(\alpha x)=(\beta\alpha)x\in Ax\), so \(Ax\) is a left ideal of \(A\). In fact, one can check that this is the smallest left ideal containing \(x\), namely the intersection of all left ideals containing \(x\), and the same argument applies to right ideals and two-sided ideals.

More generally, if we define the sum \(\mathfrak{a}+\mathfrak{b}\) of left ideals of \(A\) as the 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 it is the smallest left ideal containing both \(\mathfrak{a}\) and \(\mathfrak{b}\). Then for any elements \(x_1,\ldots, x_n\) of \(A\), the ideal

\[Ax_1+\cdots+Ax_n\]

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

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

where \(AxA\) is the finite sum of elements of the form \(\alpha x\beta\). Then the above two sets are respectively the smallest right ideal and the smallest two-sided ideal containing \(x_1,\ldots, x_n\). 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 \((x_1,\ldots, x_n)\).

On the other hand, 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 by definition always contains \(1\), if some subring is simultaneously an ideal then, as observed just above, it must be all of \(A\). Even if we weaken the notion of subring to that in \(\Rng\) and do not require the inclusion of \(1\), the condition that an ideal must be closed under multiplication by arbitrary elements is not a condition 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 The center \(Z(A)\) of a ring \(A\) is the set of elements that commute with every element 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 it 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 \(y\in A\),

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

where the element \(x\) of the center commutes with \(y\), but \(a\) need not, so \(ax\) 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 \(ax=a\not\in Z(A)\). 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 is no ideal \(\mathfrak{a}\) satisfying \(\mathfrak{m}\subsetneq\mathfrak{a}\subsetneq A\), we call \(\mathfrak{m}\) a maximal ideal.

Let any ideal \(\mathfrak{a}\subsetneq A\) of \(A\) be given. Then we can consider the collection of ideals of \(A\) that contain \(\mathfrak{a}\) and are different from \(A\) itself. 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 an ideal \(\mathfrak{a}\subsetneq A\) of a ring \(A\), there always exists a maximal ideal \(\mathfrak{m}\) of \(A\) containing \(\mathfrak{a}\).

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References

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

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