Definition of a Ring

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Ring

Definition 1 A monoid object in the symmetric monoidal category \((\Ab,\otimes, \mathbb{Z})\) is called a ring. (§Abelian Groups, ⁋Definition 11)

Following the convention for abelian groups, the operation of \(A\) is written as \(+\). In this situation, the multiplication \(\mu:A\otimes A \rightarrow A\) of a ring \(A\) is written as \(\cdot\), and when there is no danger of confusion, this is omitted and \(\alpha\cdot \beta\) is simply written as \(\alpha\beta\). Then from

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

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

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

hold. Also, \(\mu\) is associative, and

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

determines the (multiplicative) identity element \(1\) of \(A\) through the image of \(1\in \mathbb{Z}\). In other words, a set \(A\) being 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 a rng (a ring without the \(i\)). In any case, in most cases we will use a commutative ring (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 equation

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

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

2. By the result of 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, in Definition 1 we do not assume that \(0\neq 1\), but if \(0=1\) then by 1 of Proposition 2, for any \(\alpha\in A\)

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

would hold, so \(A=\{0\}\). We call such a ring the zero ring.

Ring Homomorphism

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

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

hold.

Taking these as morphisms,

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

etc. are defined. In \(\Ring\) and \(\cRing\), \(\mathbb{Z}\) is the initial object and \(\{0\}\) is the terminal object, and 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 the kernel \(\ker f\) of a ring homomorphism \(\phi\) is defined as the kernel of this group homomorphism. That is, \(\ker \phi=\phi^{-1}(0)\). Then by §Homomorphisms, ⁋Proposition 3, \(\phi\) being injective is equivalent to \(\ker \phi=\{0\}\).

Free Ring over an Abelian Group

By definition, a ring is an abelian group, and a ring homomorphism is also a homomorphism between abelian groups. That is, 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 \(\alpha_n\) of \(G^{\otimes n}\) can be written in the form

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

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, for each \(\alpha_n\), by looking at how many elements’ tensor products the constituent \(\alpha^i_{n1}\otimes\cdots \alpha^i_{nn}\) consist of, we can know which \(G^{\otimes n}\) it belongs to, so abusing notation we may 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 need to define multiplication on \(F(G)\), and 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 through the above equation the product of all elements of \(F(G)\) is defined. And we define

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

By the coherence theorem we 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 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 \(i:G\hookrightarrow F(G)\) gives an abelian group homomorphism \(\phi\circ i:G \rightarrow U(A)\).

Conversely, given any abelian group homomorphism \(f:G \rightarrow U(A)\), the following formula

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

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

We can check that the two functions \(\Hom_\Ring(F(G), A) \rightarrow\Hom_\Ab(G, U(A))\) and \(\Hom_\Ab(G, U(A))\rightarrow \Hom_\Ring(F(G), A)\) defined this way are inverses of each other, 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 \((S,+,-,\cdot,0,1)\) has a ring structure.

Meanwhile, the following holds.

Proposition 6 For any ring homomorphism \(\phi:A \rightarrow B\), \(\ker \phi\) is a subring of \(A\).

Proof

We have already checked that \(\ker \phi\) is a subgroup of the abelian group \((A,+,0)\), so 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 look closely at the above proof, we can see that even if only one of the two elements \(\alpha,\beta\) belongs to \(\ker \phi\), we can still check that \(\alpha\beta\) 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 simultaneously a left ideal and a right ideal, we call it a two-sided ideal.

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

It is easy to prove that the intersection of left ideals is a left ideal. Meanwhile, for any element \(x\) of \(A\), the following 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 \(A\). These are in fact the smallest left ideal containing \(x\), that is, they coincide with 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 following set

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

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

\[Ax_1+\cdots+Ax_n\]

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

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

and these are respectively the smallest right ideal and two-sided ideals 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 these are sometimes collectively written as \((x_1,\ldots, x_n)\).

Meanwhile, for any left ideal \(\mathfrak{a}\), we have \(1\in\mathfrak{a}\) if and only if \(\mathfrak{a}=A\). Therefore for \(\mathfrak{a}\subsetneq A\) we must necessarily have \(1\not\in\mathfrak{a}\).

Definition 8 For a ring \(A\) and an ideal \(\mathfrak{m}\), if there exists no ideal \(\mathfrak{a}\) satisfying \(\mathfrak{m}\subsetneq\mathfrak{a}\subsetneq A\), then we call \(\mathfrak{m}\) a maximal ideal.

Let any ideal \(\mathfrak{a}\) of \(A\) be given. Then we can consider the collection of ideals of \(A\) containing \(\mathfrak{a}\) and different from \(A\) itself. Then this collection is an inductive set, so it has a maximal element. ([Set Theory] §Axiom of Choice, ⁋Theorem 4) It is not difficult to see that this maximal element is a maximal ideal. That is, we obtain the following.

Theorem 9 (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.