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Algebra-II: Companion notes

Abhay Soman

Section 5.2 Chain conditions and Principal Ideal Domains

Definition 5.2.1.

A ring is said to satisfy ascending chain condition (a.c.c) if every ascending chain of (left or right or two-sided) ideals stabilize, i.e, if \(\mfa_1\subseteq\mfa_2\subseteq\cdots\mfa_n\subseteq\cdots\) is a chain of (left or right or two-sided) ideals then there exists \(N\in\N\) such that \(\mfa_N=\mfa_i\) for any \(i\geq N\text{.}\)

Definition 5.2.2.

A ring is said to satisfy descending chain condition (d.c.c) if every descending chain of (left or right or two-sided) ideals stabilize, i.e, if \(\mfa_1\supseteq\mfa_2\supseteq\cdots\mfa_n\supseteq\cdots\) is a chain of (left or right or two-sided) ideals then there exists \(N\in\N\) such that \(\mfa_N=\mfa_i\) for any \(i\geq N\text{.}\)

Definition 5.2.3. (Principal ideal domain).

An integral domain in which every ideal is principal, i.e., every ideal can be generated by an element, is said to be a principal ideal domain.

Convention 5.2.4.

We abbreviate a principal ideal domain by PID.
Let \(A\) be a PID and let \(\mfa_1\subseteq\mfa_2\subseteq\cdots\) be an ascending chain of ideals in \(A\text{.}\) Thus, \(\mfa=\bigcup_i\mfa_i\) is an ideal of \(A\text{.}\) Suppose that \(\mfa_i=(a_i)\) for each \(i\text{.}\) Since \(A\) is PID, there exists \(a\in\mfa\) such that \(\mfa=(a)\text{.}\) Suppose that \(a\in\mfa_n\text{.}\) We claim that \(\mfa_n=\mfa_i\) for all \(i\geq n\text{.}\) Indeed, since \((a)=\bigcup_i\mfa_i\) we have \(\mfa_i\subseteq(a)\text{.}\) Conversely, as \(a\in\mfa_n\subseteq\mfa_i\) for every \(i\geq n\text{,}\) the ideal generated by \(a\text{,}\) \((a)\subseteq\mfa_i\) for every \(i\geq n\text{.}\)
Let \(A\) be a PID and let \(p\in A\) be an irreducible. Suppose that \(I\) is an ideal containing \((p)\text{.}\) Since \(A\) is PID there exists \(a\in I\) such that \(I=(a)\text{.}\) Hence \(a\mid p\text{.}\) Since \(p\) is irreducible either \(a\sim p\) or \(a\) is a unit. In other words, either \((p)=(a)\) or \((a)=A\text{.}\)
By Lemma 3.3.3, an ideal generated by an irreducible element in a PID will be a prime ideal. Thus we get the following.
Let \(F\) be a field and let \(F[x,y]\) be the polynomial ring over \(F\) in two variables. Note that \(F[x,y]=F[x][y]\) hence it is UFD (we prove this result in Corollary 7.2.3). We show that the ideal generated by \(x\) and \(y\text{,}\) \((x,y)\) is not principal. Suppose that \(f\in F[x,y]\) is such that \((x,y)=(f)\text{.}\) By Corollary 4.4.2,
\begin{equation*} x=q_1f\quad\text{and}\quad y=q_2f\quad\text{for}\;q_i\in F[x,y]. \end{equation*}
Note that \(x\) considered as an element in the polynomial ring over F[x] in the variable \(y\) is a constant polynomial. Thus, \(f\) when viewed as a polynomial over \(F[x]\) in the variable \(y\) must have degree zero. Similarly the degree of \(f\) as a polynomial in \(F[y][x]\) is zero. Hence \(f\in F\) and thus \((x,y)=F[x,y]\text{.}\) However, \(x+1\not\in (x,y)\text{.}\) Therefore, \((x,y)\) is not a principal ideal.
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