Primary Ideals and Decompositions (15.2.28,31-32)

Dummit and Foote Abstract Algebra, section 15.2, exercises 28, 31-32:

28. Prove that each of the following rings have infinitely many minimal prime ideals, and that $(0)$ is not the intersection of any finite number of these (so $(0)$ does not have a primary decomposition in these rings):
(a) The infinite direct product ring $ℤ/2ℤ×ℤ/2ℤ×...$
(b) $k[x_1,x_2,...]/(x_1x_2,x_3x_4,...)$
31-32. Let $Q_1$ and $Q_2$ be $P$-primary ideals in a commutative ring with identity $R$. Show that $Q_1∩Q_2$ is also a $P$-primary ideal, and prove that if $P$ is a maximal ideal, then $Q_1Q_2$ and $Q_1+Q_2$ are also $P$-primary ideals.

Proof: 28. (a) Note that since every element of $ℤ/2ℤ×ℤ/2ℤ×...$ is multiplicatively idempotent, it follows that the quotient of $ℤ/2ℤ×ℤ/2ℤ×...$ with any of its prime ideals results in an integral domain with two elements, $1$ and $0$. Therefore, every prime ideal in this ring is a maximal ideal, and in particular is minimal. If we let $I_n$ denote the ideal of elements whose $n^{\text{th}}$ coordinate is zero, then we easily observe an infinite collection of prime/maximal ideals. Now, we shall prove that given any ideal $I$ with the property that it contains an element $x$ which has infinitely many nonzero coordinates, then $P∩I$ also has this property when $P$ is a prime ideal; inductively, this will prove particularly that any finite intersection of prime ideals is nonzero. To wit, write $x=y+z$ for $yz=0$ and each of $y,z$ having infinitely many nonzero coordinates. Then since $yz∈P$ we must have at least one of $y,z∈P$, say $y∈P$. Then since $y=yx$ we also have $y∈I$, hence $y∈P∩I$ as claimed.

(b) For any prime ideal $P⊆R=k[x_1,x_2,...]/(x_1x_2,x_3x_4,...)$, since $\overline{x_1}\overline{x_2}=0$ we must have $\{\overline{x_{2n-1}},\overline{x_{2n}}\}∩P≠ø$ for each $n∈ℕ$. Conversely, if $S⊆ℕ$ is such that exactly one of $2n-1$ or $2n$ is in $S$ for each $n∈ℕ$, and $P_S=(x_i~|~i∈S)⊆k[x_1,x_2,...]$, then $k[x_1,x_2,...]/P_S≅R/\overline{P_S}$ is an integral domain and hence $\overline{P_S}$ is a prime ideal. It follows that the minimal prime ideals of $R$ are exactly $\overline{P_S}$ for such sets $S$, of which there are (uncountably) infinitely many. Now, let $P_{S_1},...,P_{S_n}$ be any finite collection of such ideals. For each $1≤i≤n$ let $α_i∈\{2i-1,2i\}$ be such that $\overline{x_{α_i}}∈P_{S_i}$. Then since provided each $m_1,m_2...$ is monomial, for polynomials $p∈(m_1,m_2,...)$ if and only if each monomial term of $p$ is divisible by some $m_j$, we see $β=x_{α_1}x_{α_2}...x_{α_n}∉(x_1x_2,x_3x_4,...)$ while $β∈P_{S_1}∩...∩P_{S_n}$, so that $\overline{P_{S_1}}∩...∩\overline{P_{S_n}}$ is nonzero.

31-32. Observe: $$\text{rad }(Q_1Q_2)=\text{rad }(Q_1∩Q_2)=(\text{rad }Q_1)∩(\text{rad }Q_2)=P∩P=P$$ $$\text{rad }(Q_1+Q_2)=\text{rad }(\text{rad }Q_1+\text{rad }Q_2)=\text{rad }(P+P)=\text{rad }P=P$$ Therefore it suffices to prove $Q_1∩Q_2$ is a primary ideal, and that when $P$ is maximal, so too are $Q_1Q_2$ and $Q_1+Q_2$.

Suppose $xy∈Q_1∩Q_2$ where $y∉Q_1∩Q_2$, say $y∉Q_1$. Then $x^n∈Q_1⊆P=\text{rad }Q_2$ for some $n$, and therefore $x^{nm}∈Q_1∩Q_2$ for some further $m$.

Now suppose $P$ is maximal. Now $Q_1Q_2$ and $Q_1+Q_2$ are primary by Proposition 19.4.$~\square$