Spring 2026: Math 831 Daily Update

We began the semester with some basic terminology, giving proofs and examples where appropriate.

Terminology.

1. (i) \(I\subseteq R\) is an ideal if \(a, b\in I\) implies \(a+b\in I\) and \(r\in R\) implies \(ra\in I\).
2. (ii) \(P\subseteq R\) is a prime ideal if \(P\) is an ideal and \(ab\in I\) implies \(a\in I\) or \(b\in I\). We showed that if \(R := \mathbb{Q}[x,y]\), then \(P := \{fx\ |\ f\in R\}\) is a prime ideal.
3. (iii) \(M\subseteq R\) is a maximal ideal if \(M\) is a proper ideal not contained in any other proper ideal. We showed that for \(R\) as in (ii), the set of polynomials with zero constant term is a maximal ideal. Using Zorn's lemma, we proved
   Proposition. Maximal ideals exist in \(R\). In particular, every proper ideal of \(R\) is contained in a maximal ideal.
4. (iv) \(0 \not = a\) is a zerodivisor if there exists \(0\not = b\) such that \(ab = 0\). Otherwise \(a\) is a non-zerodivisor.
5. (v) \(R\) is an integral domain if every non-zero element in \(R\) is a non-zerodivisor.
6. (vi) \(u\in R\) is a unit if there exists \(v\in R\) such that \(uv = 1\).
7. (vii) \(R\) is a field if every non-zero element of \(R\) is a unit.
8. (viii) The sum of ideals \(I, J\) is the ideal \(I+J := \{i+j\ | \ i\in I\ \text{and}\ j\in J\}\).
9. (ix) For the subset \(X\subseteq R\), \(\langle X\rangle\), the ideal generated by \(X\) consists of all finite \(R\)-linear combinations of the elements of \(X\). This was observed to the same as the intersection of all ideals containing \(X\). Special cases were \(\langle a\rangle\), a principal ideal, and \(X = \{x_1,\ldots, x_n\}\) so that \(\langle X\rangle := \{r_1x_1+\cdots + r_nx_n\ |\ r_i\in R\}\).
10. (x) Given ideals \(I, J\subseteq R\), the product of \(I\) and \(J\) is the ideal \(IJ := \langle \{ij\ |\ i\in I\ \text{and}\ j \in J\}\rangle\).
11. (xi) For \(n\geq 1\), we have \(I^n\), the \(n\)th power \(I\), which is the ideal consisting of all finite sums of \(n\)-fold products of elements from \(I\).
12. (xii) A map \(\phi :R\to S\) between rings is a ring homomorphism if \(\phi (a+b) = \phi (a)+\phi (b)\) and \(\phi (ab) = \phi (a)\phi(b)\), for all \(a, b\in R\).
13. (xiii) Given an ideal \(I\subseteq R\), the factor ring \(R/I\) consists of all cosets of the form \(a+I\), with \(a\in R\), subject to:
    • \(a+I = b+I\) if and only if \(a-b\in I\).
    • \((a+I)+(b+I) := (a+b)+I\).
    • \((a+I)\cdot (b+I) := ab+I\).

We then proved that every maximal ideal in \(R\) is a prime ideal and gave an example where the converse fails, namely \(P = \langle x\rangle \subseteq \mathbb{Q}[x,y]\). From this we initiated our discussion of the process of localization, by giving the following definition.

Examples. \(S\) is a multiplicatively closed set in each of the following cases:

1. (i) \(S := \{1, f, f^2, f^3, \ldots \}\), for any \(f\in R\) such that \(f^n\not = 0\), for all \(n\geq 0\).
2. (ii) \(S\) is the set of non-zerodivisors in \(R\).
3. (iii) \(S := R\backslash P\), for \(P\) a prime ideal of \(R\).

We ended class with the following:

We began class by defining the nilradical \(\sqrt{I}\) of an ideal \(I\subseteq R\) as the set of elements \(r\in R\) such that \(r^n\in I\), for some \(n\geq 1\). We also noted that \(\sqrt{(0)}\) is called the nilradical of \(R\) and consists of the nilpotent elements in \(R\). We then gave a proof of the following corollary to Krull's Lemma,

For a multiplicatively closed set \(S\subseteq R\) we then defined an equivalence relation on \(R\times S\) as follows: \((r_1,s_1)\sim (r_2,s_2)\) if and only if there exists \(s_3\in S\) such that \(s_3(r_1s_2-r_2s_1) = 0\). After showing that \(\sim\) is an equivalence relation, we wrote the class of \((r,s)\) as a fraction \(\frac{r}{s}\) and showed that the expected operations

are well defined and make \(R_S\) into a commutative ring whose zero element is \(\frac{0}{1} = \frac{0}{s}\), for any \(s\in S\) and whose multiplicative identity is \(\frac{1}{1} = \frac{s}{s}\), for all \(s\in S\).

We then noted the following:

1. (i) The natural map \(\phi: R\to R_S\) taking \(r\) to \(\frac{r}{1}\) is a ring homomorphism.
2. (ii) \(\frac{r}{s} = \frac{0}{1}\) in \(R_S\) if and only if \(s'r = 0\) in \(R\) for some \(s'\in S\). In particular, the kernel of \(\phi\) is the set \(\{r\in R\ |\ sr = 0, \text{for some}\ s\in S\}\).
3. (iii) \(\frac{s}{1}\) is a unit in \(R_S\) for all \(s\in S\).

We then discussed the correspondence between the ideals of \(R\) and those of \(R_S\). For an ideal \(I\subseteq R,\) we noted \(I_S := \{\frac{i}{s}\ |\ i\in I\ \text{and}\ s\in S\}\) is an ideal of \(R_S\). Similarly, we noted that \(\phi^{-1}(J)\) is an ideal of \(R\) for every ideal \(J\subseteq R_S\). Later in the lecture we further noted that \(\phi^{-1}(J)\) has the property that \(sr\in \phi^{-1}(J)\), for some \(s\in S\) implies \(r\in \phi^{-1}(J)\).

We then showed that this correspondence was not quite a perfect one-to-one correspondence between the ideals of \(R\) disjoint from \(S\) and the ideals in \(R_S\). On the one hand we do have \(J = \phi^{-1}(J)_S\), for all ideals \(J\subseteq R_S\), which shows that any ideal \(J\subseteq R_S\) is \(I_S\), for some ideal \(I\subseteq R\). On the other hand, it is not the case that \(\phi^{-1}(I_S) = I\). In fact, we saw that \(\phi^{-1}(I_S) = \{r\in R\ |\ sr\in I, \ \text{for some}\ s\in S\}\), which often strictly contains \(I\). For example, if \(R = \mathbb{Z}\), \(I = \langle 6\rangle\), and \(S = \{1,2,2^2,2^3,2^4,\ldots\}\), then \(\phi^{-1}(I_S) = \langle 3\rangle\) which is strictly larger than \(I\). We ended class by proving the following theorem which shows the correspondence in question is one-to-one for prime ideals.

We began class by proving the following local-global principle:

We then began our study of unique factorization domains, UFDs. We discussed with details, the following

Preliminaries. Let \(R\) be an integral domain with quotient field \(K\).

1. (i) Cancellation holds in \(R\), i.e., if \(a\not = 0\) and \(ab = ac\), then \(b = c\).
2. (ii) \(u\in R\) is a unit if there exists \(v\in R\) such that \(uv = 1\).
3. (iii) If \(a\not = 0\), we say a divides b, written \(a\mid b\), if there exists \(r\in R\) such that \(b = ra\). This happens if and only if \(\langle b\rangle \subseteq \langle a\rangle\).
4. (iv) \(a\mid b\) and \(b\mid a\) if and only if \(\langle a\rangle = \langle b\rangle\) if and only if \(b= ua\), for some unit \(u\in R\). In this case we say that \(a\) and \(b\) are associates.
5. (v) The non-zero, non-unit \(q\in R\) is said to be irreducible if whenever \(q = ab\), for \(a,b\in R\), then either \(a\) or \(b\) is a unit.
6. (vi) The non-zero, non-unit \(p\in R\) is said to be prime if whenever \(p\mid ab\), then \(p\mid a\) or \(p\mid b\). By induction, if \(p\) is prime and \(p\) divides \(a_1\cdots a_n\), then \(p\) divides \(a_i\), for some \(i\).
7. (vii) For \(p, q, u\in R\) with \(u\) a unit, \(q\) is irreducible if and only if \(qu\) is irreducible and \(p\) is prime if and only if \(pu\) is prime.
8. (viii) For \(p, q\in R\), \(q\) is irreducible if and only if \(\langle q\rangle\) is maximal among principal ideals and \(p\) is prime if and only if \(\langle p\rangle\) is a prime ideal.
9. (ix) Every prime element is irreducible, but not conversely.
10. (x) In the ring \(R = \mathbb{Q}[t^2, t^3]\), with \(t\) an indeterminate, the element \(t^2\) is irreducible, but not prime. Another classic example: In the ring \(\mathbb{Z}[\sqrt{-5}]\), \(3, 2\pm \sqrt{-5}\) are irreducible elements that are not prime.

We noted that for the classical example from algebraic number theory, with \(R = \mathbb{Z}[\sqrt{-5}]\) can be shown that \(3\cdot 3 = 9 = (2+\sqrt{-5})\cdot (2-\sqrt{-5})\) gives two distinct factorizations into irreducible elements in the ring \(\mathbb{Z}[\sqrt{-5}]\), showing that the proposition fails for irreducible elements. It is precisely this failure that prevents most of the rings in commutative algebra, algebraic geometry and algebraic number theory from being unique factorization domains. This led to the following

A crucial component in the proof of the theorem was that if \(R\) is a UFD then every irreducible element is a prime element. Thus, for a UFD, an element is prime if and only if it is irreducible.

We ended class by presenting, but not verifying, the following examples to illustrate how subtle the UFD property can be.

Examples. 1. If \(R\) is a UFD, then \(R[x]\) is a UFD. Thus, the polynomial rings \(\mathbb{Z}[x_1, \ldots, x_n]\) and \(K[x_1, \ldots, x_n]\) are UFDs, for \(K\) any field.

2. \(\mathbb{C}[x,y]/\langle x^2+y^2-1\rangle\) is a UFD, but \(\mathbb{R}[x,y]/\langle x^2+y^2-1\rangle\) is not a UFD.

3. \(\mathbb{C}[x,y,z]/\langle x^2+y^2+z^2-1\rangle\) is not a UFD, but \(\mathbb{R}[x,y,z]/\langle x^2+y^2+z^2-1\rangle\) is a UFD.

4. \(\mathbb{C}[x,y,z]/\langle x^2+y^3+z^5\rangle\) is a UFD, but \(\mathbb{C}[x,y,z]/\langle x^2+y^3+z^4\rangle\) is not a UFD.

We continued our discussion of UFDs with the following sequence of results.

With this Definition+Proposition in hand, we were able to prove the following important characterization of UFDs.

An immediate and easy corollary was that any PID is a UFD: After all, any prime ideal is principal, necessarily generated by a prime element.

We then turned to the following proposition, the second part of which is known as Nagata's Lemma.

The surprising condition (iii) in Theorem A played a key role in the proof of this proposition. We then stated and proved the following major theorem.

The proof of the theorem relied on Nagata's Lemma and the following facts which we did not prove: (i) For \(a\in R\), \(a\mid f(x)\) in \(R[x]\) if and only if, in \(R\), \(a\) divides every coefficient of \(f(x)\) and (ii) If \(p\in R\) is a prime element, then \(p\) is still a prime element in \(R[x]\). The proof of the theorem then followed using Nagata's Lemma: Let \(S\) be the non-zero elements of \(R\). Thus every element of \(S\) is a product of prime elements. By the facts just stated, every element of \(S\) is a product of prime elements in \(R[x]\), thus \(R[x]\) is a UFD, if \(R[x]_S = R_S[x]\) is a UFD. But \(R_S = K\), the quotient field of \(R\) and \(R_S[x] = K[x]\) is a PID, and hence a UFD.

As corollaries we noted that the polynomial rings \(\mathbb{Z}[x_1, \ldots, x_n]\) and \(K[x_1, \ldots, x_n]\) are UFDs, for \(K\) any field.