# 4.3: Quotient Rings: New Rings from Old

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- 82473

In this section, we'll seek to answer the questions:

- How can we use ideals to build new rings out of old?
- What sorts of ideals allow us to build domains? Fields?
- What is the First Isomorphism Theorem?

If the only rings that existed were polynomial rings, familiar systems of numbers like \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}\text{,}\) and matrix rings, there would still be enough to justify the defining the concept of a ring and exploring its properties. However, these are not the only rings that exist. In this section, we explore a way of building new rings from old by means of ideals. To better understand these new rings, we will also define two new classes of ideals: prime ideals, and maximal ideals. We end by briefly connecting these rings to a familiar problem from high school algebra.

## 4.3.1Congruence modulo I

The major concept of this section is the notion of congruence modulo \(I\text{.}\) One can reasonably think of this idea as a generalization of congruence modulo \(m\) in \(\mathbb{Z}\text{.}\)

Let \(R\) be a ring and \(I\) an ideal of \(R\text{.}\) Then elements \(a,b\in R\) are said to be **congruent modulo**\(I\) if \(b-a\in I\text{.}\) If this is the case, we write \(a + I = b + I\text{.}\)

Determine (with brief justification) whether \(a + I = b + I\) in the following rings \(R\text{.}\)

- \(a = 9\text{,}\) \(b = 3\text{,}\) \(I = \langle 6\rangle\text{,}\) \(R = \mathbb{Z}\)
- \(a = 10\text{,}\) \(b = 4\text{,}\) \(I = \langle 7\rangle\text{,}\) \(R = \mathbb{Z}\)
- \(a = 9\text{,}\) \(b = 3\text{,}\) \(I = \langle 6\rangle\text{,}\) \(R = \mathbb{Z}[x]\)
- \(a = x^2+x-2\text{,}\) \(b = x-1\text{,}\) \(I = \langle x+1\rangle\text{,}\) \(R = \mathbb{Q}[x]\)
- (
*Challenge.*) \(a = x^3\text{,}\) \(b = x^2+2x\text{,}\) \(I = \langle y-x^2, y-x-2\rangle\text{,}\) \(R = \mathbb{Q}[x,y]\)

Given a ring \(R\text{,}\) ideal \(I\text{,}\) and \(a\in R\text{,}\) when is it the case that \(a + I = 0 + I = I\text{?}\)

Observe that if \(b-a \in I\text{,}\) then there is some \(x\in I\) such that \(b-a = x\text{,}\) and so \(b = a+x\text{.}\)

As was the case in \(\mathbb{Z}_m\text{,}\) congruence modulo \(I\) is an equivalence relation.

Let \(R\) be a ring and \(I\) an ideal of \(R\text{.}\) Then congruence modulo \(I\) is an equivalence relation on \(R\text{.}\)

The set of equivalence classes under this relation is denoted \(R/I\text{.}\) What is more, this is not merely a set of equivalence classes. As the next two theorems demonstrate, this set possesses two algebraic operations that extend naturally from those of \(R\text{.}\)

Let \(R\) be a ring and \(I\) an ideal of \(R\text{.}\) If \(a,b,c,d\in R\) such that \(a+I = b+I\) and \(c+I = d+I\text{,}\) then \((a+c) + I = (b+d) + I\text{.}\)

Let \(R\) be a ring and \(I\) an ideal of \(R\text{.}\) If \(a,b,c,d\in R\) such that \(a+I = b+I\) and \(c+I = d+I\text{,}\) then \(ac + I = bd + I\text{.}\)

The previous two theorems together show that addition and multiplication on the set \(R/I\) is well-defined. As these operations are built on the operations of \(R\text{,}\) it will likely not surprise you to learn that the usual axioms defining a ring also hold.

Let \(R\) be a commutative ring with identity \(1_R\) and \(I\) an ideal of \(R\text{.}\) The set of equivalence classes modulo \(I\text{,}\) denoted \(R/I\text{,}\) is a commutative ring with identity \(1_R + I\) under the operations of addition modulo \(I\) and multiplication modulo \(I\) defined in Theorem 4.3.2 and Theorem 4.3.3 .

Thus, given a ring \(R\) and ideal \(I\) of \(R\text{,}\) we may build a new ring \(R/I\text{.}\) In Subsection 4.3.2, we will explore the question of when \(R/I\) possesses some of the properties we've previously explored, e.g., when is \(R/I\) a domain? A field? First, we conclude with two explorations. The first gives us a sense of what these rings can look like. The second connects quotient rings to *solution sets* of polynomial equations.

Consider the ring \(R=\mathbb{Z}_2[x]\) and the ideals \(I = \langle x^2-1\rangle\) and \(J = \langle x^3 -x -1\rangle\text{.}\)

- List the elements of \(R/I\) and \(R/J\text{.}\)
- What happens to \(x^2\) in \(R\) when you pass to the quotient ring \(R/I\text{?}\) How about \(x^3\) as you pass from \(R\) to \(R/J\text{?}\)
- In view of your answer to the previous question, how does \(x\) behave as you “mod out” by \(I\) and \(J\text{?}\)
- Build addition and multiplication tables for each of \(R/I\) and \(R/J\text{.}\)

One of the most useful connections made in high school algebra is the connection between a function \(f\) (in particular, a polynomial function) and its *graph*. We may extend this notion to ideals via the concept of a *zero set* as follows.

Let \(F\) be a field and \(R = F[x,y]\) with \(I\subseteq R\) a nonzero ideal. We define the zero set of \(I\text{,}\) denoted \(Z(I)\text{,}\) as the set of all points \((a,b)\in F^2\) for which \(f(a,b)=0\) for all \(f\in I\text{.}\)

- Suppose \(I = \langle f_1, f_2, \ldots, f_n\rangle\text{.}\) Prove that \((a,b)\in Z(I)\) if and only if \(f_j(a,b) = 0\) for each \(j\in \{1,\ldots, n\}\text{.}\) Thus, \(Z(I)\) can be determined entirely by examining the generators of \(I\text{.}\)
- Describe \(Z(I)\) given \(I = \langle y-x^2\rangle \text{.}\)
- (Challenge) Given \(I = \langle y-x^2\rangle\) and \(J = \langle y-x-2\rangle\text{,}\) describe \(Z(I+J)\) and \(Z(I\cap J)\text{.}\)
- Given \(I=\langle y-x^2\rangle\text{,}\) describe the relationship between the variables \(x\) and \(y\) in the quotient \(R/I\text{.}\) In what way have we restricted our polynomial “inputs” to the parabola \(y = x^2\text{?}\)

## 4.3.2Prime and Maximal Ideals

In this section, we continue our exploration of quotient rings by looking more closely at properties of ideals. We focus on particular properties of ideals that ensure that the quotient \(R/I\) is either a domain or a field.

Let \(R\) be commutative with identity and \(P\subsetneq R\) a nonzero ideal. We say \(P\) is **prime** if whenever \(a,b\in R\) such that \(ab\in P\text{,}\) we have \(a\in P\) or \(b\in P\text{.}\)

Let \(R\) be a domain and \(p\in R\) be prime. Then \(\langle p\rangle\) is a prime ideal.

Which of the following ideals are prime?

- \(\langle 9\rangle\) in \(\mathbb{Z}\)
- \(\langle 11\rangle\) in \(\mathbb{Z}\)
- \(\langle x^2+1\rangle\) in \(\mathbb{R}[x]\)
- \(\langle x^2-1\rangle\) in \(\mathbb{R}[x]\)
- \(\langle x^2-5x+6, x^4+2x^3-10x^2+5x-2\rangle\) in \(\mathbb{R}[x]\)

It is this precise condition that guarantees that the resulting quotient is a domain.

Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Then \(I\) is prime if and only if \(R/I\) is an integral domain.

We now consider another important class of ideals: the maximal ideals.

Let \(R\) be commutative with identity and let \(M\subsetneq R\) be a nonzero ideal. We say that \(M\) is a **maximal ideal**** **if no proper ideal of \(R\) properly contains \(M\text{.}\) That is, if \(J\) is an ideal satisfying \(M\subseteq J\subseteq R\text{,}\) either \(J=M\) or \(J=R\text{.}\)

In other words, an ideal \(M\ne R\) is maximal if no “larger” ideal (with respect to inclusion) properly contains it. As we will see later, rings can have many maximal ideals.

It is a fact that any ring \(R\) with \(0_R\ne 1_R\) has a maximal ideal. This follows from *Zorn's Lemma*; a rigorous exploration of Zorn's Lemma lies outside of the scope of this text, but suffice it to say that Zorn's Lemma is incredibly useful in all areas of algebra for proving existence theorems. For example, a proof that every vector space has a basis relies on Zorn's Lemma.

Rings with only one maximal ideal are said to be *local rings*, and are actively studied in modern research in commutative algebra (the study of commutative rings and their properties).

The next two results demonstrate that the maximality of \(I\) is precisely the condition that guarantees that \(R/I\) is a field.

Let \(R\) be commutative with identity and \(M\) a maximal ideal of \(R\text{.}\) Let \(x\in R\setminus M\text{,}\) and set \(J = \{xr+y: r\in R, \ y\in M\}\text{.}\) Then \(M\subsetneq J\text{,}\) and thus there exist \(r'\in R\text{,}\) \(y'\in M\) such that \(1 = xr'+y'\text{.}\)

Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Then \(I\) is maximal if and only if \(R/I\) is a field.

**Hint**-
For the forward direction, apply the previous lemma to construct an inverse for \(x+I\) given any \(x\in R\setminus I\text{.}\)

Every maximal ideal is prime.

In general, the converse is not true (see the Challenge below). However, it holds in sufficiently nice rings.

In a principal ideal domain, every prime ideal is maximal.

Describe the prime and maximal ideals of \(\mathbb{Z}\) and \(\mathbb{Q}[x]\text{.}\)

**Hint**-
For which ideals \(I\) is \(\mathbb{Z}/I\) a domain? A field? Similarly for \(\mathbb{Q}[x]\text{.}\) Or, use Theorem 4.3.9 .

Find a commutative ring with identity, \(R\text{,}\) and a nonmaximal prime ideal \(P\) of \(R\text{.}\)

## 4.3.3Homomorphisms and Quotient Rings

As quotient rings provide fertile soil for building new examples of rings, it should not surprise us to find that homomorphisms interact with quotient rings in interesting and useful ways. Chief among them are the *isomorphism theorems*. In this subsection, we focus primarily on the First Isomorphism Theorem.

We have seen that any homomorphism \(\varphi : R\to S\) gives rise to an ideal of \(R\text{,}\) namely \(\ker\varphi\text{.}\) Our next theorem demonstrates that, given a commutative ring with identity \(R\text{,}\) every ideal is the kernel of some homomorphism defined on \(R\text{.}\)

Let \(R\) be commutative with identity and \(I\) an ideal of \(R\text{.}\) Define \(\varphi: R\to R/I\) by \(\varphi(r) = r+I\text{.}\) Then \(\varphi\) is a homomorphism with \(\ker\varphi = I\text{.}\)

In what follows, we work toward a proof of the First Isomorphism Theorem for Rings.

Throughout, let \(R\) and \(S\) be commutative rings with identity, and let \(\varphi : R\to S\) be a homomorphism. Recall that \(\text{im } \varphi = \{s\in S : \varphi(r) = s\text{ for some } r\in R\}\text{.}\)

Define \(f: R/\ker \varphi \to \text{im } \varphi\) by \(f(r+\ker \varphi) = \varphi(r)\text{.}\)

Using the notation from above, \(f\) is a well-defined function.

Using the notation above, \(f\) is a homomorphism.

Using the notation above, \(f\) is one-to-one.

Using the notation above, \(f\) is onto.

We thus obtain:

Let \(\varphi : R\to S\) be a homomorphism of commutative rings. Then \(R/\ker \varphi \cong \text{im } \varphi\text{.}\)

In particular, if \(\varphi : R\to S\) is onto, \(R/\ker \varphi \cong S\text{.}\)

The First Isomorphism Theorem gives a useful way of establishing an isomorphism between a quotient ring \(R/I\) and another ring \(S\text{:}\) find an onto homomorphism \(R\to S\) with kernel \(I\text{.}\)

We have the following isomorphisms of rings.

- \(\displaystyle \mathbb{Z}/\langle m\rangle \cong \mathbb{Z}_m\)
- \(\displaystyle \mathbb{Q}[x]/\langle x-5\rangle \cong \mathbb{Q}\)
- \(\displaystyle \mathbb{R}[x]/\langle x^2+1\rangle \cong \mathbb{C}\)

Let \(R = \mathbb{Z}_6\) and define \(\varphi : \mathbb{Z}_6 \to \mathbb{Z}_2\) by \(\varphi(\overline{x}) = \overline{x}\text{.}\) That is, \(\varphi\) sends an equivalence class \(\overline{x}\in \mathbb{Z}_6\) represented by \(x\in \mathbb{Z}\) to the equivalence class represented by \(x\) in \(\mathbb{Z}_2\text{.}\)

- Show that \(\varphi\) is a well-defined function.
- Prove that \(\varphi\) is a homomorphism.
- Is \(\varphi\) onto? Justify.
- Compute \(\ker\varphi\) (that is, list the elements in the set). Is \(\varphi\) one-to-one?
- Without appealing to the definition, is \(\ker\varphi\) prime? Maximal? Explain.