Basic Algebra I.4 (Ideals and Ring Quotients)

Our first naïve attempt is to take a ring quotient R/S for a subring S of R.  First, (S, +) is a subgroup of (R, +) so we can denote every coset R/S by x+S for some x in R. Naturally we’ll want our 1+S to be the identity element in R/S, which immediately gives us a problem: since 1 is in S, 1+S = 0+S and we have 1=0 in R/S. This means r’ = r’×1 = r’×0 = 0 for any r’ in R/S which is ridiculous.

This forces us to abandon our approach. Instead we look at subrngs of R. Let I be any subrng. Once again, every coset R/I is represented by x+I for some x. In order for multiplication to be well-defined in R/I, we need the following condition:

x+I = x'+I,\,\, y+I = y'+I \implies xy+I = x'y'+I.,

i.e. x - x' \in I, \,\, y-y' \in I \implies xy - x'y' \in I.

But xy – x’y’ = x(yy’) + (xx’)y’ and a moment of reflection tells us that our condition above is equivalent to the following:

if x \in I, y \in R, then xy, yx \in R.

Hence we define an ideal to be a subset I of R such that:

  • I contains 0;
  • if xy lies in I, then x+y lies in I;
  • if x lies in I and y lies in R, then xy and yx lie in I.

Sometimes, one also says I is a two-sided ideal; the reason for this will become apparent when we think of ideals as a special case of modules.

Note: from the conditions above, it follows that I is a subgroup of (R, +); indeed to show that I is closed under subtraction, write x-y = x + (-1)y and apply the last property.

Examples of ideals

  1. In Z, the only ideals are nZ where n is a non-negative integer.
  2. In a field K, the only ideals are {0} and K. [ Indeed, if I is an ideal containing some non-zero element x, then x has an inverse y; since xy = 1, we see that I contains 1 and hence every element of K.  ]
  3. In a commutative ring R, for any x\in R, we can take the set \left<x\right> := \{ xy : y\in R\}  of all multiples of x. Clearly, this is non-empty, closed under addition, and closed under multiplication by any element of R.
  4. Recall that the set of upper-triangular real n × n matrices forms a ring. The set of strictly upper-triangular matrices (whose diagonal entries are 0) forms an ideal.
  5. The ring M(n, R) of n × n real matrices forms a ring with no non-trivial ideals. [ Sketch of proof : let x be a non-zero matrix in ideal I. Left- and right-multiply by appropriate matrices such that the resulting matrix has only 1 non-zero entry…. ]

In example 5, the ring in question has no ideals other than 0 and itself. We call such a ring a simple ring (just like a person who has no ideals is a simple person).

 

 

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