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Handout Monday 11/03
Today we will explore ideals and their quotient rings, as well as what happens when a subring is not an ideal.
Recall that an ideal \(J\) is a subring of a ring \(R\) such that for all \(r \in R\) and \(a \in J\text{,}\) \(ra \in J\text{.}\) That is, it is a subring that is closed under multiplication by elements outside of the subring (with an element in the subring).
We already saw some examples of ideals. In \(\Z\text{,}\) we can look at \(\langle n \rangle\text{,}\) the set of all multiples of some constant integer \(n\text{.}\) In fact, any “ideal generated by an element” will be an ideal, for any commutative ring. Let’s check this.
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Prove that \(\langle a \rangle = \{ra \st r \in R\}\) is an ideal.
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Let \(b, c \in \langle a \rangle\text{.}\) That means \(b = r_1a\) and \(c = r_2a\text{.}\) Then \(b+c = r_1a+r_2a = (r_1+r_2)a \in \langle a\rangle\text{.}\) Similarly, \(bc = r_1ar_2a = r_1r_2aa \in \langle a \rangle\text{.}\)
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If \(b \in \langle a \rangle\text{.}\) So \(b = ra\) for some \(r\text{.}\) But then \(-b = -ra\) so \(-b \in \langle a \rangle\)
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Finally, let \(b \in \langle a \rangle\) and \(r \in R\text{.}\) We know \(b = r_2 a\text{.}\) But then \(rb = rr_2a \in \langle a \rangle\text{.}\) So \(\langle a \rangle\) is closed under external products.
Let’s look at the ring \(\Q\) (in fact, a field). What is \(\langle \frac{1}{2}\rangle\text{?}\) All the multiples of \(\frac{1}{2}\) is definitely an ideal. But what all is in it? We get every element of \(\Q\text{,}\) since this includes \(1 = 2 \cdot \frac{1}{2}\)
Are there other subrings of \(\Q\text{?}\) Well, the integers are definitely a ring, and every integer is a rational numbers, so \(\Z\) is a subring of \(\Q\text{.}\) But is \(\Z\) an ideal in \(\Q\text{?}\) Let’s see: if we take an element in \(\Z\) and multiply it by an element of the parent ring, do we land back in \(\Z\text{?}\) Definitely not!
So \(\Z\) is not an ideal. Let’s see what happens when we form the cosets of \(\Z\) in \(\Q\text{.}\) What is \(\Q/\Z\text{?}\)
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We definitely get cosets: \(a + \Z\) for any \(a \in \Q\text{.}\) So we have things like \(\frac{1}{2} + \Z\) and \(\frac{1}{3}+\Z\text{.}\) What are these really? How many cosets are there?
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Can we perform coset addition here? Yes, since \(\Q\) is abelian, we know that \(\Z\) is a normal subgroup, so coset addition is well defined. Let’s make sure this works with the two cosets we picked above.
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What about coset multiplication? This is where we expect a problem. What is \((\frac{1}{2}+\Z)(\frac{1}{3}+ \Z)\text{?}\) We need this to be \(\frac{1}{6}+\Z\text{.}\) So what’s the problem?
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To show an operation is not well-defined, we should pick different names of the coset and see if we get a problem. Another name for these cosets is \(\frac{3}{2}+\Z\) and \(\frac{4}{3}+\Z\text{.}\) What is their product?
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It would need to be \(\frac{12}{6}+\Z = 2+\Z = \Z\text{.}\)
Another non-ideal example: Look at the ring \(\Q[x]\) of all polynomials with integer coefficients. \(\Q\) is a subring, thinking of it as the constant polynomials. Find an example of the cosets failing under multiplication.
Both of these examples are still interesting as quotient groups though. What does \(\Q/\Z\) look like? It is really a circle of rational numbers. \(\Q[x]/\Q\) contains a coset for each polynomial, but two polynomials that differ only by a constant are equivalent (in the same coset).
