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Basic Ring Theory

Table of Contents

1. Definition

The study of rings generalizes the study of number systems. The archetypical number systems include our favorites: the integers, the rationals, the reals, and the complex numbers. But not the natural numbers (because we want to include subtraction).

A Ring consists of

  • a set \(R\)

equipped with

  • addition \(+\colon R\times R\to R\)
  • negation \(-\colon R\to R\)
  • a zero-like object \(0\in R\)
  • multiplication \(-\cdot-\colon R\times R\to R\)
  • a one-like object \(1\in R\)

such that

  • Addition is associative \((x + y) + z = x + (y + z)\)
  • Addition is commutative \(x + y = y + x\)
  • Zero is the unit for addition: \(x + 0 = 0 + x = x\)
  • Negation gives the additive inverse: \(x + (-x) = 0\)
  • Multiplication is associative \(x\cdot(y\cdot z) = (x\cdot y)\cdot z\)
  • One is the unit for multiplication \(x\cdot1=1\cdot x=x\)
  • Multiplication distributes over addition \((x + y)z = xz + yz\)

If, in addition, multiplication is commutative, then we call the ring a Commutative Ring.

If we abandon the existence of a multiplicative identity, then we have a non-unital ring.

There are probably a few more properties I'm forgetting about.

(Internalization, category theoretic definition.) We could define a "ring object" as a monoid object in the category Ab of Abelian groups. (For this reason, we could have defined a ring as consisting of an Abelian group equipped with multiplication.) This gives us a sense of the important properties: we need to have an underlying Abelian group. We can give up, e.g., commutativity for multiplication (we could even give up associativity!), but we really want to have commutativity for addition.

2. Examples

The integer \(\ZZ\) is the epitome of a ring, as are the rationals \(\QQ\), and the reals \(\RR\).

Let \(R\) be a commutative ring, then the collection of polynomials in a single indeterminate \(z\) with coefficients in \(R\) is a ring…called the Polynomial Ring over \(R\), denoted \(R[z]\).

We can generalize the polynomial ring in a number of ways: we could consider rational polynomials \(R(z) = \{f(z)/g(z)\mid f,g\in R[z], g\neq0\}\); or formal power series \(R[{[z]}]\); or the formal Laurent series \(R((z))\) which intuitively is \(R[z^{-1}] + R[{[z]}]\).

But not the formal distributions over \(R\), i.e., \(R[{[z,z^{-1}]}]\) because multiplication is not consistent (this pops up in vertex operator algebras).

Let \(R\) be a ring, and \(n\in\NN\) a natural number greater than zero. Then the set of \(n\)-by-\(n\) matrices with entries in \(R\), denoted \(M(n,R)\), forms another ring under matrix addition and matrix multiplication.

This is an important example, because for \(n\gt1\) the ring is noncommutative.

Let \(X\) be some non-empty set, and \(R\) a ring. Then the function space \(\hom(X,R)\) of \(R\)-valued functions on \(X\) forms a ring. Addition is defined pointwise \(f+g\) is the function \((f+g)(x) = f(x)+g(x)\), as are multiplication \((f\times g)(x)=f(x)g(x)\). The additive identity is the constant function \(Z(x) = 0\) for all \(x\in X\), and the multiplicative identity is the function \(e(x) = 1\) for all \(x\in X\).

If \(X\) is a topological space, then continuous real-valued functions form a ring. (Continuous complex-valued functions form another ring.)

Let \(R\) be a noncommutative ring. Can we extend the notion of a polynomial ring to \(R[x]\)?

Let \(n\in\NN\) be greater than 1, \(R\) some fixed commutative ring. Is \(M(n, R[z])\) the same ring as \((M(n, R))[z]\)? What happens for noncommutative rings \(R\)?

3. Special Elements in a Ring

Let \(r\in R\) be an element in a ring. We call it Nilpotent if there exists some \(k\in\NN\) such that \(r^{k}=0\).

We could consider nilpotents as modeling "infinitesimals" of yore.

Let \(r\in R\) be an element in a ring. We call it Idempotent if \(r^{2}=r\).

4. Ring Morphisms

Let \(R\) and \(S\) be rings. A Ring Morphism is a function of the underlying sets \(f\colon R\to S\) such that

  1. it preserves addition \(f(a+b) = f(a) + f(b)\)
  2. it preserves multiplication \(f(ab) = f(a)f(b)\)
  3. it preserves "one" \(f(1_{R})=1_{S}\).

As a consequence of preserving addition, we have \(f(0+0)=f(0)+f(0)=f(0)\) which implies \(f(0)=0\). Further \(f(0-a) = f(0) + f(-a) = f(-a)\) and \(f(a-a) = f(a) + f(-a) = 0\) implies \(f(-a)=-f(a)\).

Let \(R\) be a ring. Then we always have a morphism \(f\colon\ZZ\to R\) defined by \(f(0)=0\), \(f(1)=1\), \(f(2)=f(1)+f(1)\), and more generally for \(n\in\NN\) we have

\begin{equation} f(n) = \sum^{n}_{j=1}f(1). \end{equation}

Sometimes we abuse notation and write \(f(n)=nf(1)\).

Let \(R\) be a ring, \(r\in R\) some fixed element. Then we have a natural morphism \(e_{r}\colon R[z]\to R\) by evaluating a polynomial at \(z=r\), i.e., \(e_{r}(p(z)) = p(r)\).

For a ring \(R\) and the polynomial ring \(R[z]\), the inclusion mapping \(R\to R[z]\) mapping \(r\in R\) to \(r\in R[z]\) is a ring morphism.

But the function mapping a polynomial to its roots in \(R\), \(\rho\colon R[z]\to\mathcal{P}(R)\), is not a ring morphism. How can we see this? Well, the codomain isn't necessarily a ring (what's the additive inverse? What's addition?), but \(\rho(p(z) + q(z))\) is not necessarily mapped to \(\rho(p(z)) + \rho(q(z))\).

Let \(R\) and \(S\) be rings, we call the ring morphism \(f\colon R\to S\) a Ring Isomorphism (or just isomorphism) if there is a morphism \(g\colon S\to R\) such that \(g\circ f=\mathrm{id}_{R}\) and \(f\circ g=\mathrm{id}_{S}\) (i.e., an inverse morphism exists).

A ring isomorphism, like all isomorphisms, tells us when two rings are "the same".

5. Subrings, Adjoining Elements

Let \(R\) be a ring, \(S\subset R\) some subset of \(R\). We call \(S\) a Subring of \(R\) if

  1. \(S\) is a subgroup of \(R\) under addition,
  2. \(S\) is closed under multiplication from \(R\),
  3. \(1\in S\).

The integers \(\ZZ\) are a subring of the rationals \(\QQ\); the rationals are a subring of the reals \(\RR\); the reals are again a subring of the complex numbers \(\CC\).

A non-example: the modular ring \(\ZZ/(24)\) of hours in a day, this is not a subring of \(\ZZ\). The operators are distinct.

Let \(S\) be a subring of a commutative ring \(R\). We can Adjoin an Element \(r\in R\) to \(S\) — denoted \(S[r]\) — by taking the subring of formal polynomials \(S[z]\subset R[z]\) and the evaluation morphism \(e_{r}\colon R[z]\to R\) restricted to the subring \(S[z]\subset R[z]\) gives us \(S[r] = e_{r}(S[z])\subset e_{r}(R[z])\).

We can consider \(\ZZ\subset\CC\) as a subring, then adjoining \(\I=\sqrt{-1}\) to \(\ZZ\) gives us the Gaussian Integers \(\ZZ[\I]\). We basically treat \(\I\) as a variable, then replace \(\I^{2}\) by \(-1\) (and commutativity lets us factor \(\I^{n+2}=\I^{2}\I^{n}\) to iterate this heuristic). An arbitrary Gaussian integer looks like \(a+\I b\) for arbitrary integers \(a,b\in\ZZ\). Multiplication is defined as

\begin{equation} (a + \I b)(x + \I y) = (ax - by) + \I(ay + bx). \end{equation}

Addition is defined component-wise.

We can generalize this example to any square-free number \(d\in\NN\), then \(\ZZ[\sqrt{-d}]\) is another ring adjoined with a new element. (The example of \(\ZZ[\sqrt{-5}]\) is thrown around a lot because in that ring, \(6=2\cdot3=(1-\sqrt{-5})(1+\sqrt{-5})\) has two distinct factorizations.)

6. Ideals

Let \(f\colon R\to S\) be a ring morphism, \(0_{S}\in S\) be the additive identity ("zero") of \(S\). We define the Kernel of \(f\) to be the subset

\begin{equation} \ker(f) = \{r\in R\mid f(r) = 0_{S}\}. \end{equation}

Observe that the kernel of a ring morphism is closed under addition — if \(x,y\in\ker(f)\), then \(f(x+y) = f(x) + f(y) = 0+0 = 0\) — and multiplication. Moreover, for any \(r\in R\) and for any \(x\in\ker(f)\), we see that \(f(xr)=f(x)f(r)=0\), and \(f(rx)=f(r)f(x)=0\); i.e., that the kernel is closed under multiplication by elements of the domain on the right and left, respectively.

This idea is captured by the following definition:

Let \(R\) be a ring.

  1. A Left Ideal in \(R\) is a subset \(I_{L}\subset R\) is such that (a) \(0\in I_{L}\), (b) \(x+y\in I_{L}\) whenever both \(x\in I_{L}\) and \(y\in I_{L}\), (c) \(ry\in I_{L}\) whenever \(y\in I_{L}\) for any \(r\in R\).
  2. A Right Ideal in \(R\) is a subset \(I_{R}\subset R\) is such that (a) \(0\in I_{R}\), (b) \(x+y\in I_{R}\) whenever both \(x\in I_{R}\) and \(y\in I_{R}\), (c) \(xr\in I_{R}\) whenever \(x\in I_{R}\) for any \(r\in R\).
  3. A Two-Sided Ideal in \(R\) is a subset \(I\subset R\) is such that (a) \(0\in I\), (b) \(x+y\in I\) whenever both \(x\in I\) and \(y\in I\), (c) \(xy\in I_{L}\) whenever \(x\in I\) or \(y\in I\).

Further, we call the ideal Proper if the ideal is a proper subse of the ring.

For commutative rings, these three coincide, and we just speak of "Ideals". It's for noncommutative rings we need to distinguish them. Whenever results depend on the kind of ideal, we make note of it. Otherwise, we say "ideal" with the understanding it may mean any of the three kinds.

Let \(I\) be an ideal of \(R\). Then \(I\) is a subgroup of the underlying Abelian group of \(R\) (equipped with addition).

Let \(R\subset R'\) be a proper subring, and \(q\in R'\) an element not in \(R\). Then \(R[q]\) is a ring obtained by taking \(R[z]\subset R'[z]\), then applying the evaluation morphism \(e_{q}\colon R'[z]\to R'\) restricted to \(R[z]\). That is to say, \(e_{q}(R[z])\subseteq R'\).

It's not uncommon to see things like, if \(r\in R\), then \(rR = \{rx\mid x\in R\}\).

Let \(R\) be a ring, \(S\subset R\) some nonempty subset. We describe the Ideal Generated by \(S\) to be the sums of elements of the form \(xsy\) where \(s\in S\) and \(x,y\in R\).

For commutative rings, we can use commutativity to rewrite \(xsy=xys\) and thus \((S) = RS\).

Let \(I\) and \(J\) be ideals of \(R\). Notation:

  1. Then we denote the ideal generated by \(I\cup J\) as \(I+J\).
  2. We also denote \(IJ\) is the ideal generated by elements of the form \(ab\) for \(a\in I\) and \(b\in J\).

(Jacobson) Let \(I\), \(J\), \(K\) be ideals of \(R\). Does distributivity hold for "ideal arithmetic"? I.e., is \((I+J)K\) equal to \(IK + JK\)?

Let \(R\) be a ring, \(I\) an ideal of \(R\). Prove or find a counter-example: \(IR\subseteq I\). Is this true for noncommutative rings?

7. References

  • Manuel Bronstein, Symbolic Integration 1: Transcendental Functions. Springer, second ed., 1996. Chapter 1 particularly reviews polynomial rings.
  • Nathan Jacobson, Basic Algebra I. Chapter 2.
  • Serge Lang, Algebra. Springer, third ed., GTM., 2002. Chapter 2.
  • James McIvor, Lecture Notes on Ring Theory. UC Berkeley, Math 113, Summer 2014.

Last Updated 2021-06-01 Tue 10:00.