Section Week 11 (11/03-11/8)
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).
Subsection Wednesday 11/5
Today we will explore what it means for two groups to be “the same”. Start with the Same or different? activity.
If there is time, specify what it means for a function to be an isomorphism.
Subsection Friday 11/7
We explored more about isomorphisms building on what we started the previous class. In particular, we thought about how you can build an isomorphism by making sure elements go to other elements that have the same order. That is, you make a choice and use that to “generate” the rest of the isomorphism.
