# Ideals - Sharp: Steps in Commutative Algebra - Chapter 2

• Dec 7th 2012, 05:12 PM
Bernhard
Ideals - Sharp: Steps in Commutative Algebra - Chapter 2
I am reading Sharp: Steps in Commutative Algebra, Chapter 2: Ideals. (see attached pages 18-19 for Sharp's definitions etc)

On page 19 Sharpe gives the following exercise

2.4 EXERCISE

Let X be an indeterminate and consider the ring $\mathbb{Q}$ [X] of polynomials in X with co-efficients in $\mathbb{Q}$

Give

(i) an example of a subring of $\mathbb{Q}$ [X] which is not an ideal of $\mathbb{Q}$ [X]

and

(ii) an example of an ideal of $\mathbb{Q}$ [X] which is not a subring of $\mathbb{Q}$ [X]

I would appreciate help with this exercise.

Peter
• Dec 8th 2012, 11:24 AM
Deveno
Re: Ideals - Sharp: Steps in Commutative Algebra - Chapter 2
(i) what is the difference between a subring, and an ideal?

a) both are additive subgroups, so addition doesn't tell us anything useful.

b) subrings are closed under multiplication, ideals are as well...but: ideals have an additional property: they "absorb" elements of the parent ring R by multiplication.

to get a better understanding of ideals, it helps to know a bit of history: ideals were modelled after "ideal numbers". we can extend the notion of "ordinary" integers to "complex integers" (or gaussian integers...that is, the ring Z[i]). but Z has a property Z[i] does not: unique factorization (for example, in Z[i], we have 3 = (1)(3) = (2+i)(2-i)). so there was a search for things which could serve as a suitable replacement for "primes" in "algebraic number rings" to preserved unique factorization.

without getting overly technical, it would realized that unique factorization could be preserved, if "numbers" were replaced by "ideal numbers". at first, "ideal" numbers were just conceived of as "hypothetical subdivisors of primes" (more technically, enlarging our ring so it contains "more divisors"). over time, it was realized that the "ideal numbers" could be replaced by the ideals (in our current sense) they generated. so for a much larger class of rings, we had unique factorization not of elements as products of prime elements, but any ideal as products of prime ideals.

ideals (particularly principal ideals, those generated by a single element) capture what we really mean by "divisibility": when we say a divides b, what we mean is: b is in the ideal generated by a. equivalently:

there is a surjective ring homomorphism φ:R-->R/(a) for which φ(b) = 0.

this last view is "intuitively helpful", it shows that the elements of an ideal can capture the following properties of 0 in any ring:

0 + 0 = 0
r0 = 0r = 0 <---this is the "absorbtion" property that distinguishes an ideal I from a mere subring...it "absorbs" outside elements.

********

so...on to specifics with R = Q[X]. consider S = Q[X2], consisting of all polynomials in Q[X] with only even powers of X (including X0 = 1). this is clearly an additive subgroup of Q[X], and also closed under multiplication. so it's a subring of Q[X]. but it's NOT an ideal.

for example, we have X2 in S, and certainly X is in Q[X], but X(X2) = X3 is NOT in S.

for the other problem, it depends on your definition of "subring": not that the set S = (X), consisting of all polynomials in Q[X] with constant term 0 is indeed an ideal, but it lacks one thing Q[X] has: a multiplicative identity. if one does not require that a subring contain a multiplicative identity, then every ideal is also a subring (some authors use the term rng to denote "ring without identity", so we have: an ideal is a subrng of a ring).
• Dec 8th 2012, 01:48 PM
Bernhard
Re: Ideals - Sharp: Steps in Commutative Algebra - Chapter 2
Deveno,

That post is extremely helpful and informative.

(You should be lecturing/tutoring in a University!)

Peter
• Feb 15th 2013, 01:25 AM
arulanandam
Re: Ideals - Sharp: Steps in Commutative Algebra - Chapter 2
Find the number of annihilating ideals in the Ring F1XF2XF3 where F's are Fields