Introduction to Ring Theory (3)

Ideals and Ring Quotients

Suppose I is a subgroup of (R, +). Since + is abelian, I is automatically a normal subgroup and we get the group quotient (R/I, +). One asks when we can define the product operation on R/I.

To be specific, each coset in R/I is of the form (r+I), for some r in R. We wish to define the product operation on R/I via (r+I)(s+I) := rs+I. The question arises as to whether the operation is well-defined, i.e.

If r+I = r’+I and s+I = s’+I, then is it true that rs+I = r’s’+I ?

Now, r+I = r’+I iff $r-r' \in I$ . So we need the condition:

$r-r', s-s' \in I \implies rs-r's'\in I.$

But then we have:

$rs - r's' = r\overbrace{(s-s')}^{\in I} + \overbrace{(r-r')}^{\in I}s'.$

Letting r = r’ and s’ = 0, we get the condition $s\in I \implies rs\in I$ for each r in R. Likewise, in letting s = s’ and r’ = 0, we get $r\in I \implies rs\in I$ for each s in R. Clearly, these conditions suffice to ensure that the above hold.

Definition. An (two-sided) ideal of R is an additive subgroup I of (R, +) such that $r\in R, s \in I \implies rs, sr\in I$ . Note that if R is commutative, then rs=sr so it suffices to check for $rs\in I$ .

From the above reasoning, an ideal I of R gives a ring quotient R/I which is the set of all cosets r+I, with additional and multiplication given by (r+I)+(s+I) := (r+s)+I and (r+I)(s+I) = (rs)+I. The corresponding unity is given by 1+I.

Examples

Every additive subgroup of the ring Z is of the form nZ, where n ≥ 0. It’s clear that this is an ideal. For n=0, we have Z/{0} = Z. Otherwise, n>0 and Z/nZ is precisely the ring of integers mod n, which we had denoted by Z/n.
Let R be a division ring. Certainly, {0} and R are ideals, with respective quotients R and the trivial ring. Conversely, suppose I is an ideal of R which contains x≠0. Then x is a unit so there exists a y in R such that xy=1. So $1 = xy \in I$ by definition of ideal. But this means for any r in R, $r = 1\cdot r \in I$ , i.e. I = R. This means a division ring has no non-trivial ideals. In particular, a field has no non-trivial ideal.
The ring of 2 × 2 real matrices has no non-trivial ideal too. This will follow from a later chapter on matrix rings.
If U denotes the ring of 2 × 2 upper-triangular real matrices $\begin{pmatrix} * & *\\ 0 & *\end{pmatrix}$ , then the subset I of strictly upper-triangular real matrices $\begin{pmatrix} 0 & *\\ {0}& 0\end{pmatrix}$ is an ideal since $\begin{pmatrix}0 & *\\{0}&0\end{pmatrix}$ $\begin{pmatrix}* & *\\{0}& *\end{pmatrix}$ = $\begin{pmatrix}0 & *\\{0}& 0\end{pmatrix}$ and similarly $\begin{pmatrix}* & *\\{0} & *\end{pmatrix}$ $\begin{pmatrix} 0 & * \\ {0} & 0\end{pmatrix}$ = $\begin{pmatrix}0 & *\\{0} & 0\end{pmatrix}$ . The ring quotient U/I is isomorphic to R × R by mapping the two diagonal entries (a and b) of a matrix to $(a,b)\in\mathbf{R}\times\mathbf{R}$ .
In the ring R × R, the only ideals are {(0, 0)}, {0} × R, R × {0} and R × R. This follows from the following.
1. If R and S are rings, then all ideals of R × S are of the form I × J, where I, J are respectively ideals of R, S.
2. Is it true that all subrings of R × S are of the form T × U, where T, U are respectively subrings of R, S?
Every ideal of R[x] is of the form <p(x)>, the set of multiples of p(x), where p(x)=0 or p(x) is a monic polynomial. This will follow from a later chapter on polynomial rings. This is not true for Z[x]: for example, the set $\{2a(x) + x\cdot b(x) : a(x), b(x) \in \mathbf{Z}[x]\}$ is an ideal which is not of the above form.

Operations on Ideals

The following are a list of common operations on ideals. Their proofs are easy and left as exercises.

Addition. If I and J are ideals of R, let $I+J=\{x+y : x\in I, y\in J\}$ . This is an ideal of R. In fact, it is the “smallest ideal of R containing I and J“. [ This just means (i) it contains both I and J, and (ii) any ideal containing I and J must contain it. ]
Intersection. If {I_i} is a collection of ideals of R, then so is the intersection I = ∩_iI_i.
Product. If I and J are ideals of R, let $IJ = \{x_1 y_1 + \ldots + x_n y_n : x_i\in I, y_i \in I\}$ . Then IJ is an ideal of R.

A few words about the product, since the reader may not be accustomed to it. Note that we only allow finitely many terms in the sum, although the number of terms has no upper-bound. Also, this differs from the “pairwise products” which one may be inclined to define via $S=\{xy : x\in I, y\in J\}$ .

Exercise.

Consider the ring of integers Z. As stated above, each ideal is of the form mZ, where m is a non-negative integer. Determine the ideals mZ + nZ, mZ ∩ nZ and mZ·nZ by describing them in terms of number-theoretic operations on integers.

Answer (highlight to read): [ Addition, intersection and product of ideals correspond to the lowest common multiple, greatest common divisor and product of the corresponding non-negative integers. ]

Generated Ideals

Now, since the intersection of ideals is an ideal, let’s define the generated ideal as before.

Definition. Let X be an arbitrary subset of R. The intersection of all ideals containing X is denoted by <X> and is called the ideal generated by X.

By now, you should have seen the term “generated by” many times and have a good idea what that means. Under this definition, the product of ideals IJ is simply the ideal generated by the set of pairwise products: $S=\{xy : x\in I, y\in J\}$ .

What does the ideal <X> look like? For one thing, it must contain every x of X, and hence by definition of ideals, it contains rxs for any $r, s\in R$ . Since <X> is closed under addition, it must contain all finite sums of the form $r_1 x_1 s_1 + \ldots r_n x_n s_n$ , where $x_i \in X$ and $r_i, s_i \in R$ . On the other hand, the set of all such elements clearly satisfies the conditions of being an ideal. Thus:

$\left<X\right> = \{r_1 x_1 s_1 + \ldots + r_n x_n s_n : x_i \in X, r_i, s_i \in R\}.$

Note that the x_i‘s are not necessarily distinct, since there’s no generic way to simplify $r_1 x s_1 + r_2 x s_2$ any further. However, if R is commutative, then we only need to consider:

$\left<X\right> = \{r_1 x_1 + \ldots + r_n x_n: x_i \in X, r_i\in R\}$

and the x_i‘s may be take to be distinct since $r_1 x + r_2 x = (r_1 + r_2)x$ . In particular,

for a singleton set X={x}, $\left<x\right> = \{rx : r\in R\}$ (for notational convenience, we set <x> := <{x}>) – such ideals are called principal ideals;
for two-element set X={x, y}, $\left<x,y\right> = \{rx + sy : r,s\in R\}$ ;
etc etc.

Examples (all commutative)

Let’s pick R[x]. Then the ideal <f(x)> is the set of all multiples of f(x).
Again pick R[x]. Consider the ideal I = <x²-1, x³-1>. A priori, it’s generated by two elements. Now is it principal (i.e. generated by one element)? The answer is yes. In fact, we claim I = <x-1>. Indeed: since x²-1 and x³-1 are both multiples of x-1, clearly I is contained in <x-1>. Conversely, x-1 = (x³-1) – x(x²-1) is an element of I. So <x-1> is contained in I. This completes our proof. [ Note: we’ll see later that every ideal of this ring is principal. This will have important consequences. ]
Pick Z[x]. Consider the ideal I = <2, x-1>. Is it principal? If it were, then we would have a polynomial f(x) such that <2, x-1> = <f(x)>. Since 2 is an element of <f(x)> it is a multiple of f(x). This can only happen if f(x)=-2, -1, 1 or 2, none of which works. So I is not a principal ideal in Z[x].
Pick the ring of Gaussian integers Z[i] from the previous post and take the ideal I = <8+i, 5-14i>. Is this ideal principal? Now that’s a little tricky: but it turns out it is! We claim I = <2-3i>. Indeed: since 8+i = (2-3i)(1+2i) and 5-14i = (2-3i)(4-i) are both multiples of 2-3i, I must be contained in <2-3i>. Conversely, 2-3i = 2(8+i) – i(5-14i) is an element of I, so <2-3i> is contained in I. Thus, I is principal! The reader may wonder how one can compute such relations effectively. This will be covered in a later post. [ We’ll also see that every ideal of Z[i] is principal, which marks the birth of algebraic number theory. ]