# Commutative Algebra

## UFDs, PIDs, etc

### Spring 2013 #2#algebra/qual/completed

• Define a Euclidean domain.

• Define a unique factorization domain.

• Is a Euclidean domain an UFD? Give either a proof or a counterexample with justification.

• Is a UFD a Euclidean domain? Give either a proof or a counterexample with justification.

• $$R$$ is Euclidean iff it admits a Euclidean algorithm: there is a degree function $$f: R\to {\mathbf{Z}}_{\geq 0}$$ such that for all $$a,b\in R$$, there exist $$q, r\in R$$ such that $$a = bq + r$$ where $$f(r) <f(b)$$ or $$r=0$$.

• $$R$$ is a UFD iff every $$r\in R$$ can be written as $$r = u \prod_{i=1}^n p_i$$ with $$n\geq 0$$, $$u\in R^{\times}$$, and $$p_i$$ irreducible. This is unique up to associates of the $$p_i$$ and reordering.

• Euclidean implies UFD:

• Euclidean implies PID:
• If $$I \in \operatorname{Id}(R)$$ one can use the degree function to find any $$b \in I$$ where $$f(b)$$ is minimal.
• Then $$I = \left\langle{b}\right\rangle$$, since if $$a\in I$$ one can write $$a = bq + r$$ and use that $$a-bq \in I \implies r\in I$$.
• But by minimality, we can’t have $$f(r)<f(b)$$, so $$r=0$$ and $$a \divides b$$, so $$b\in \left\langle{a}\right\rangle$$.
• PID implies UFD:
• Use that irreducible implies prime in a PID, so every $$x\in R$$ has some factorization into finitely many primes.
• Supposing $$x = u_p \prod_{i=1}^m p_i = u_q \prod_{i=1}^n q_i$$, use that $$p_1$$ divides the LHS and so $$p_1$$ divides the RHS. WLOG, $$p_1\divides q_1$$, so $$q_1 = u_1 p_1$$ for $$u\in R^{\times}$$, so $$x = u_q u_1 p_1 \prod_{i=2}^m q_i$$ by rewriting a term on the RHS.
• Note that this makes $$p_1, q_1$$ associates.
• Continuing up to $$m$$, we get \begin{align*} x &= u_p \prod_{i=1}^m p_i \\ &= u_q \prod_{i=1}^m u_i p_i \prod_{k=m+1}^n q_i \\ \implies u_p &= u_q \prod_{i=1}^m u_i \prod_{k=m+1}^n q_i \\ \tilde u &= \prod_{k=m+1}^n q_i ,\end{align*} where we’ve moved all units to the LHS. This makes $$p_i, q_i$$ associates for $$i\leq m$$.
• But primes aren’t units and the product of nontrivial primes can’t be a unit, so the right-hand side product must be empty.
• So $$m=n$$ and all $$p_i, q_i$$ are associate, QED.
• UFD does not imply Euclidean:

• It suffices to find a UFD that is not a PID.
• Take $$R \coloneqq{\mathbf{C}}[x, y]$$, which is a UFD by the usual factorization of polynomials. It is not a PID, since $$\left\langle{2, x}\right\rangle$$ is not principal.

### Fall 2017 #6#algebra/qual/completed

For a ring $$R$$, let $$U(R)$$ denote the multiplicative group of units in $$R$$. Recall that in an integral domain $$R$$, $$r \in R$$ is called irreducible if $$r$$ is not a unit in R, and the only divisors of $$r$$ have the form $$ru$$ with $$u$$ a unit in $$R$$.

We call a non-zero, non-unit $$r \in R$$ prime in $$R$$ if $$r \divides ab \implies r \divides a$$ or $$r \divides b$$. Consider the ring $$R = \{a + b \sqrt{-5}{~\mathrel{\Big\vert}~}a, b \in Z\}$$.

• Prove $$R$$ is an integral domain.

• Show $$U(R) = \{\pm1\}$$.

• Show $$3, 2 + \sqrt{-5}$$, and $$2 - \sqrt{-5}$$ are irreducible in $$R$$.

• Show 3 is not prime in $$R$$.

• Conclude $$R$$ is not a PID.

• Integral domain: $$ab=0 \implies a\neq 0 \text{ or } b\neq 0$$.
• Prime: $$p \divides ab \implies p\divides a$$ or $$b$$.
• Reducible: $$a = xy$$ where $$x, y$$ are proper divisors.
• Irreducible implies prime in a UFD.

• $$R$$ is an integral domain:
• Let $$\alpha = a + b\sqrt{-5}$$ and $$\beta = c + d \sqrt{-5}$$ and set $$\overline{\alpha}, \overline{\beta}$$ be their conjugates.
• Then \begin{align*} 0 = \alpha \beta = \alpha\overline{\alpha }\beta\overline{\beta }= (a^2-5b^2)(c^2-5d^2) \in {\mathbf{Z}} ,\end{align*} so one factor is zero.
• If $$a^2 = 5b^2$$ then $$a = \sqrt{5} b \not\in {\mathbf{Z}}$$ unless $$a=b=0$$. Otherwise, the same argument forces $$c=d=0$$.
• The units are $$\pm 1$$:
• Use that $$u\in R^{\times}\implies N(u) = \pm 1$$, and $$N(\alpha) = \alpha \overline{\alpha }= (a+b\sqrt{-5})(a-b\sqrt{-5}) = a^2 + 5b^2 = 1$$ forces $$b=0$$ and $$a=\pm 1$$.
• Irreducible elements:
• $$2, 3$$ are irreducible because if (say) $$3=xy$$ then $$N(x)N(y) = N(3) = 9$$, and if neither $$x,y$$ are units then $$N(x) = N(y) = 3$$. But $$N(a + b\sqrt{-5}) = a^2 + 5b^2$$ and $$a^2 + 5b^2 = 3$$ has no solutions. The same argument works for $$2$$.
• $$2\pm \sqrt{-5}$$ are irreducible because $$N(2 + \sqrt{-5}) = 2^2 + 5(1) = 9$$, and in fact $$N(2 - \sqrt{-5}) = 2^2 + 5(-1)^2 = 9$$. By the same argument as above, this forces irreducibility.
• $$3$$ is not prime:
• We can write $$6 = (3)(2) = (1 + \sqrt{-5})(1 - \sqrt{-5})$$, so if we assume $$3$$ is prime we get $$3\divides (1 \pm \sqrt{-5})$$.
• But writing $$(1\pm \sqrt{-5}) = 3r$$ for some $$r\in R$$ yields \begin{align*} (1 \pm \sqrt{-5}) = 3(a + b\sqrt{-5}) \implies 3a=1, 3b = \pm 1 .\end{align*} These have no solutions $$a, b\in {\mathbf{Z}}$$. $$\contradiction$$
• $$R$$ is not a PID:
• Use that irreducibles are prime in a UFD, which is not true here.

### Spring 2017 #4#algebra/qual/completed

• Let $$R$$ be an integral domain with quotient field $$F$$. Suppose that $$p(x), a(x), b(x)$$ are monic polynomials in $$F[x]$$ with $$p(x) = a(x) b(x)$$ and with $$p(x) \in R[x]$$, $$a(x)$$ not in $$R[x]$$, and both $$a(x), b(x)$$ not constant.

Prove that $$R$$ is not a UFD.

(You may assume Gauss’ lemma)

• Prove that $${\mathbf{Z}}[2\sqrt{2}]$$ is not a UFD.

Hint: let $$p(x) = x^2-2$$.

• Gauss’ lemma: for $$R$$ a UFD with fraction field $$F$$, if $$f$$ is reducible in $$F[x]$$ with $$f=pq$$ then there are $$r,s\in R$$ such that $$f = (rp)(sq)$$ reduces in $$R[x]$$.
• Corollary: $$R$$ is a UFD iff $$R[x]$$ is a UFD.

• The important assumption is $$a(x)\not\in R[x]$$, we’ll assume $$R$$ is a UFD and try to contradict this.
• Write $$f(x) = a(x)b(x)\in F[x]$$, then if $$R$$ is a UFD we have $$r,s\in F$$ such that $$f(x) = ra(x) sb(x) \in R[x]$$.
• Since $$a(x), b(x)$$ are monic and $$f=ab$$, $$f$$ is monic, and by the factorization in $$R[x]$$ we have $$rs=1$$. So $$r,s\in R^{\times}$$.
• Then using that $$ra(x)\in R[x]$$, we have $$r^{-1}ra(x) = a(x)\in R[x]$$. $$\contradiction$$

• Set $$R = {\mathbf{Z}}[2\sqrt 2], F = {\mathbf{Q}}[2\sqrt 2]$$.
• Let $$p(x) \coloneqq x^2-2 \in R[x]$$ which splits as $$p(x) = (x+ \sqrt{2} )(x - \sqrt{2} ) \coloneqq a(x) b(x) \in F[x]$$.
• Note neither $$a(x), b(x)$$ are in $$R[x]$$.
• Explicitly, every monic linear $$p\in R[x]$$ is of the form $$x + 2t\sqrt{2}$$ with $$t\in {\mathbf{Z}}$$, and $$\pm \sqrt{2} \neq 2t\sqrt{2}$$ for any $$t$$.
• So we have $$p(x) \in R[x]$$ splitting as $$p=ab$$ in $$F[x]$$ with $$a\not \in R[x]$$, so part (a) applies.

## Ideals (Prime, Maximal, Proper, Principal, etc)

### Fall 2021 #5#algebra/qual/stuck

Let $$R$$ be an algebra over $${\mathbf{C}}$$ which is finite-dimensional as a $${\mathbf{C}}{\hbox{-}}$$vector space. Recall that an ideal $$I$$ of $$R$$ can be considered as a $${\mathbf{C}}{\hbox{-}}$$subvector space of $$R$$. We define the codimension of $$I$$ in $$R$$ to be \begin{align*} \operatorname{codim}_R I \coloneqq \dim_{{\mathbf{C}}} R - \dim_{{\mathbf{C}}} I ,\end{align*} the difference between the dimension of $$R$$ as a $${\mathbf{C}}{\hbox{-}}$$vector space, $$\dim_{{\mathbf{C}}} R$$, and the dimension of $$I$$ as a $${\mathbf{C}}{\hbox{-}}$$vector space, $$\dim_{\mathbf{C}}I$$.

• Show that any maximal ideal $$m \subset R$$ has codimension 1 .

• Suppose that $$\operatorname{dim}_{{\mathbf{C}}} R=2$$. Show that there exists a surjective homomorphism of $${\mathbf{C}}{\hbox{-}}$$algebras from the polynomial ring $${\mathbf{C}}[t]$$ to $$R$$.

• Classify such algebras $$R$$ for which $$\dim_{{\mathbf{C}}} R=2$$, and list their maximal ideals.

(DZG): my impression is that this is an unusually difficult problem, or was something specifically covered in this year’s qual class.

Part a: Since $$I$$ is proper, we have $$\operatorname{codim}_R I \geq 1$$ since $$\operatorname{codim}_R I = 0 \implies I = R$$ since $$I\leq R$$ is a vector subspace of the same dimension. We also have $$\operatorname{codim}_R I \leq \dim_{\mathbf{C}}R$$, and noting that $$\operatorname{codim}_R I = \dim_{\mathbf{C}}R \iff \dim_{\mathbf{C}}I = \dim_{\mathbf{C}}R$$ and if $$I$$ is maximal it is necessarily proper, we in fact have $$\operatorname{codim}_R I < \dim_{\mathbf{C}}R$$, so \begin{align*} 1 \leq \operatorname{codim}_R I \leq \dim_{\mathbf{C}}R - 1 .\end{align*} Now if $$\operatorname{codim}_R I \geq 2$$, then $$\dim_{\mathbf{C}}I \leq \dim_{\mathbf{C}}R - 2$$. Choosing a basis $$\left\{{v_1,\cdots, v_n}\right\}$$ for $$R$$ as a $${\mathbf{C}}{\hbox{-}}$$vector space induces a basis $$\left\{{v_1,\cdots, v_k}\right\}$$ on $$I$$ for some $$k\leq n-2$$. But then $$I' \coloneqq\left\langle{v_1,\cdots, v_k, v_{k+1}}\right\rangle$$ is a proper $${\mathbf{C}}{\hbox{-}}$$vector subspace of $$R$$ containing $$I$$, contradicting maximality of $$I$$. So $$\operatorname{codim}_R I < 2$$, forcing $$\operatorname{codim}_R I = 1$$.

Part b: Choose a vector space basis $$\left\{{v_1, v_2}\right\}$$ for $$R$$ and define a map \begin{align*} \phi: {\mathbf{C}}[t] &\to R \\ 1 & \mapsto v_1 \\ t & \mapsto v_2 ,\end{align*} extended by linearity.

Part c: ???

### Fall 2013 #3#algebra/qual/completed

• Define prime ideal, give an example of a nontrivial ideal in the ring $${\mathbf{Z}}$$ that is not prime, and prove that it is not prime.

• Define maximal ideal, give an example of a nontrivial maximal ideal in $${\mathbf{Z}}$$ and prove that it is maximal.

• $${\mathfrak{p}}$$ is prime iff $$xy\in {\mathfrak{p}}\implies x\in {\mathfrak{p}}$$ or $$y\in {\mathfrak{p}}$$.
• An ideal $$I{~\trianglelefteq~}{\mathbf{Z}}$$ that is not prime: $$I \coloneqq 8{\mathbf{Z}}$$.
• For example, $$2\cdot 4\in 8{\mathbf{Z}}$$ but neither 2 nor 4 is a multiple of 8.
• $${\mathfrak{m}}$$ is maximal iff whenever $$I\supseteq{\mathfrak{m}}$$ is an ideal in $$R$$, then either $$I={\mathfrak{m}}$$ or $$I = R$$.
• A maximal ideal in $${\mathbf{Z}}$$: $$p{\mathbf{Z}}$$. This is because primes are maximal in a PID and $$p{\mathbf{Z}}$$ is a prime ideal. Alternatively, “to contain is to divide” holds for Dedekind domains, so $$m{\mathbf{Z}}\supseteq p{\mathbf{Z}}\iff m\divides p$$, which forces $$m=1,p$$, so either $$m{\mathbf{Z}}= p{\mathbf{Z}}$$ or $$m{\mathbf{Z}}= {\mathbf{Z}}$$.

### Fall 2014 #8#algebra/qual/work

Let $$R$$ be a nonzero commutative ring without unit such that $$R$$ does not contain a proper maximal ideal. Prove that for all $$x\in R$$, the ideal $$xR$$ is proper.

You may assume the axiom of choice.

### Fall 2014 #7#algebra/qual/completed

Give a careful proof that $${\mathbf{C}}[x, y]$$ is not a PID.

• If $$R[x]$$ is a PID, then $$R$$ is a field (not explicitly used).
• In $$P \coloneqq R[x_1, \cdots, x_n]$$, there are degree functions $$\deg_{x_n}: P\to {\mathbf{Z}}_{\geq 0}$$.

• The claim is that $$I \coloneqq\left\langle{x, y}\right\rangle$$ is not principal.
• Toward a contradiction, if so, then $$\left\langle{x, y}\right\rangle = \left\langle{f}\right\rangle$$.
• So write $$x = fg$$ for some $$g\in {\mathbf{C}}[x, y]$$, then
• $$\deg_x(x) = 1$$, so $$\deg_x(fg) = 1$$ which forces $$\deg_x(f) \leq 1$$.
• $$\deg_y(y) = 1$$, so $$\deg_y(fg) = 1$$ which forces $$\deg_y(f) \leq 1$$.
• So $$f(x, y) = ax + by + c$$ for some $$a,b,c\in {\mathbf{C}}$$.
• $$\deg_x(y) = 0$$ and thus $$\deg_x(fg) = 0$$, forcing $$a=0$$
• $$\deg_y(x) = 0$$ and thus $$\deg_y(fg) = 0$$, forcing $$b=0$$
• So $$f(x, y) = c \in {\mathbf{C}}$$.
• But $${\mathbf{C}}[x]$$ is a field, so $$c$$ is a unit in $${\mathbf{C}}$$ and thus $${\mathbf{C}}[x, y]$$, so $$\left\langle{f}\right\rangle = \left\langle{c}\right\rangle = {\mathbf{C}}[x, y]$$.
• This is a contradiction, since $$1\not\in \left\langle{x, y}\right\rangle$$:
• Every element in $$\alpha(x, y) \in\left\langle{x, y}\right\rangle$$ is of the form $$\alpha(x, y) = xp(x, y) + yq(x, y)$$.
• But $$\deg_x(\alpha) \geq 1, \deg_y(\alpha)\geq 1$$, while $$\deg_x(1) = \deg_y(1) = 0$$.
• So $$\left\langle{x, y}\right\rangle \neq {\mathbf{C}}[x, y]$$.
• Alternatively, $$\left\langle{x, y}\right\rangle$$ is proper since $${\mathbf{C}}[x, y] / \left\langle{x, y}\right\rangle \cong {\mathbf{C}}\neq {\mathbf{C}}[x, y]$$.

### Spring 2019 #6#algebra/qual/completed

Let $$R$$ be a commutative ring with 1.

Recall that $$x \in R$$ is nilpotent iff $$xn = 0$$ for some positive integer $$n$$.

• Show that every proper ideal of $$R$$ is contained within a maximal ideal.

• Let $$J(R)$$ denote the intersection of all maximal ideals of $$R$$. Show that $$x \in J(R) \iff 1 + rx$$ is a unit for all $$r \in R$$.

• Suppose now that $$R$$ is finite. Show that in this case $$J(R)$$ consists precisely of the nilpotent elements in R.

• Definitions: \begin{align*} N(R) &\coloneqq\left\{{x\in R {~\mathrel{\Big\vert}~}x^n = 0 \text{ for some } n}\right\} \\ J(R) &\coloneqq\cap_{{\mathfrak{m}}\in \operatorname{mSpec}} {\mathfrak{m}} .\end{align*}

• Zorn’s lemma: if $$P$$ is a poset in which every chain has an upper bound, $$P$$ contains a maximal element.

Define the set of proper ideals \begin{align*} S = \left\{{J {~\mathrel{\Big\vert}~}I \subseteq J < R}\right\} ,\end{align*}

which is a poset under set inclusion.

Given a chain $$J_1 \subseteq \cdots$$, there is an upper bound $$J \coloneqq\cup J_i$$, so Zorn’s lemma applies.

$$\implies$$:

• We will show that $$x\in J(R) \implies 1+x \in R^{\times}$$, from which the result follows by letting $$x=rx$$.

• Let $$x\in J(R)$$, so it is in every maximal ideal, and suppose toward a contradiction that $$1+x$$ is not a unit.

• Then consider $$I = \left\langle{1+x}\right\rangle {~\trianglelefteq~}R$$. Since $$1+x$$ is not a unit, we can’t write $$s(1+x) = 1$$ for any $$s\in R$$, and so $$1 \not\in I$$ and $$I\neq R$$

• So $$I < R$$ is proper and thus contained in some maximal proper ideal $$\mathfrak{m} < R$$ by part (1), and so we have $$1+x \in \mathfrak{m}$$. Since $$x\in J(R)$$, $$x\in \mathfrak{m}$$ as well.

• But then $$(1+x) - x = 1 \in \mathfrak{m}$$ which forces $$\mathfrak{m} = R$$.

$$\impliedby$$

• Fix $$x\in R$$, and suppose $$1+rx$$ is a unit for all $$r\in R$$.

• Suppose towards a contradiction that there is a maximal ideal $$\mathfrak{m}$$ such that $$x\not \in \mathfrak{m}$$ and thus $$x\not\in J(R)$$.

• Consider \begin{align*} M' \coloneqq\left\{{rx + m {~\mathrel{\Big\vert}~}r\in R,~ m\in M}\right\} .\end{align*}

• Since $$\mathfrak{m}$$ was maximal, $$\mathfrak{m} \subsetneq M'$$ and so $$M' = R$$.

• So every element in $$R$$ can be written as $$rx + m$$ for some $$r\in R, m\in M$$. But $$1\in R$$, so we have \begin{align*} 1 = rx + m .\end{align*}

• So let $$s = -r$$ and write $$1 = sx - m$$, and so $$m = 1 + sx$$.

• Since $$s\in R$$ by assumption $$1+sx$$ is a unit and thus $$m \in \mathfrak{m}$$ is a unit, a contradiction.

• So $$x\in \mathfrak{m}$$ for every $$\mathfrak{m}$$ and thus $$x\in J(R)$$.

$$\mathfrak N(R) \subseteq J(R)$$:

• Use the fact $$x\in \mathfrak N(R) \implies x^n = 0 \implies 1 + rx$$ is a unit $$\iff x\in J(R)$$ by (b): \begin{align*} \sum_{k=1}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1- (-x)} = (1+x)^{-1} .\end{align*}

$$J(R) \subseteq \mathfrak N(R)$$:

• Let $$x \in J(R) \setminus \mathfrak N(R)$$.

• Since $$R$$ is finite, $$x^m = x$$ for some $$m > 0$$.

• Without loss of generality, we can suppose $$x^2 = x$$ by replacing $$x^m$$ with $$x^{2m}$$.

• If $$1-x$$ is not a unit, then $$\left\langle{1-x}\right\rangle$$ is a nontrivial proper ideal, which by (a) is contained in some maximal ideal $${\mathfrak{m}}$$. But then $$x\in {\mathfrak{m}}$$ and $$1-x \in {\mathfrak{m}}\implies x + (1-x) = 1 \in {\mathfrak{m}}$$, a contradiction.

• So $$1-x$$ is a unit, so let $$u = (1-x)^{-1}$$.

• Then \begin{align*} (1-x)x &= x - x^2 = x - x = 0 \\ &\implies u (1-x)x = x = 0 \\ &\implies x=0 .\end{align*}

### Spring 2018 #8#algebra/qual/completed

Let $$R = C[0, 1]$$ be the ring of continuous real-valued functions on the interval $$[0, 1]$$. Let I be an ideal of $$R$$.

• Show that if $$f \in I, a \in [0, 1]$$ are such that $$f (a) \neq 0$$, then there exists $$g \in I$$ such that $$g(x) \geq 0$$ for all $$x \in [0, 1]$$, and $$g(x) > 0$$ for all $$x$$ in some open neighborhood of $$a$$.

• If $$I \neq R$$, show that the set $$Z(I) = \{x \in [0, 1] {~\mathrel{\Big\vert}~}f(x) = 0 \text{ for all } f \in I\}$$ is nonempty.

• Show that if $$I$$ is maximal, then there exists $$x_0 \in [0, 1]$$ such that $$I = \{ f \in R {~\mathrel{\Big\vert}~}f (x_0 ) = 0\}$$.

Cool problem, but pretty specific topological tricks needed.

• Suppose $$c\coloneqq f(a)\neq 0$$, noting that $$c$$ may not be positive.
• By continuity, pick $${\varepsilon}$$ small enough so that $${\left\lvert {x-a} \right\rvert}<{\varepsilon}\implies {\left\lvert {f(x) - f(a)} \right\rvert} < c/2$$. Since we’re on the interval, we have $$f(x) \in (f(a) - c/2, f(a) + c/2) = (c/2, 3c/2)$$ which is a ball of radius $$c/2$$ about $$c$$, which thus doesn’t intersect $$0$$.
• So $$f(x) \neq 0$$ on this ball, and $$g \coloneqq f^2 > 0$$ on it. Note that ideals are closed under products, so $$g\in I$$
• Moreover $$f^2(x) \geq 0$$ since squares are non-negative, so $$g\geq 0$$ on $$[0, 1]$$.

• By contrapositive, suppose $$V(I)= \emptyset$$, we’ll show $$I$$ contains a unit and thus $$I=R$$.
• For each fixed $$x\in [0, 1]$$, since $$V(I)$$ is empty there is some $$f_x$$ such that $$f_x(x) \neq 0$$.
• By (a), there is some $$g_x$$ with $$g_x(x) > 0$$ on a neighborhood $$U_x\ni x$$ and $$g_x \geq 0$$ everywhere.
• Ranging over all $$x$$ yields a collection $$\left\{{(g_x, U_x) {~\mathrel{\Big\vert}~}x\in [0, 1]}\right\}$$ where $$\left\{{U_x}\right\}\rightrightarrows[0, 1]$$.
• By compactness there is a finite subcover, yielding a finite collection $$\left\{{(g_k, U_k)}\right\}_{k=1}^n$$ for some $$n$$.
• Define the candidate unit as \begin{align*} G(x) \coloneqq{1\over \sum_{k=1}^n g_k(x)} .\end{align*}
• This is well-defined: fix an $$x$$, then the denominator is zero at $$x$$ iff $$g_k(x) = 0$$ for all $$k$$. But since the $$U_k$$ form an open cover, $$x\in U_\ell$$ for some $$\ell$$, and $$g_\ell > 0$$ on $$U_\ell$$.
• Since ideals are closed under sums, $$H\coloneqq{1\over G} \coloneqq\sum g_k \in I$$. But $$H$$ is clearly a unit since $$HG = \operatorname{id}$$.

• If $$I{~\trianglelefteq~}R$$ is maximal, $$I\neq R$$, and so by (b) we have $$V(I) \neq \emptyset$$.
• So there is some $$x_0\in[0, 1]$$ with $$f(x_0) = 0$$ for all $$f\in I$$.
• Define $${\mathfrak{m}}_{x_0} \coloneqq\left\{{f\in R {~\mathrel{\Big\vert}~}f(x_0) = 0}\right\}$$, which is clearly an ideal.
• This is a proper ideal, since constant nonzero functions are continuous and thus in $$R$$, not not $${\mathfrak{m}}_{x_0}$$.
• We thus have $$I \subseteq {\mathfrak{m}}_{x_0}$$, and by maximality they are equal.
\todo[inline]{I'm not super convinced by c!}


## Zero Divisors and Nilpotents

### Spring 2014 #5#algebra/qual/completed

Let $$R$$ be a commutative ring and $$a\in R$$. Prove that $$a$$ is not nilpotent $$\iff$$ there exists a commutative ring $$S$$ and a ring homomorphism $$\phi: R\to S$$ such that $$\phi(a)$$ is a unit.

Note: by definition, $$a$$ is nilpotent $$\iff$$ there is a natural number $$n$$ such that $$a^n = 0$$.

$$\not A\implies \not B$$:

• Suppose $$a$$ is nilpotent, so $$a^m = 0_R$$, and suppose $$\phi: R\to S$$ is a ring morphism.
• Ring morphisms send zero to zero, so $$0_S = \phi(0_R) = \phi(a^m) = \phi(a)^m$$ and $$\phi(a)$$ is nilpotent.
• But nontrivial rings can’t contain nilpotent units: if $$u$$ is a unit and $$ut= 1$$ with $$u^k=0$$, then $$1 = 1^k = (ut)^k = u^k t^k=0$$ and $$R=0$$.

$$A\implies B$$:

• If $$a$$ is not nilpotent, localize at the infinite multiplicative subset $$A \coloneqq\left\{{1, a, a^2, \cdots}\right\}$$ to obtain $R \left[ { \scriptstyle { {A}^{-1}} } \right]$. Since $$0\not\in A$$, this is not the zero ring.
• By the universal property, there is a map $\phi: R\to R \left[ { \scriptstyle { {A}^{-1}} } \right]$, and the claim is that $$\phi(a)$$ is a unit in $R \left[ { \scriptstyle { {A}^{-1}} } \right]$.
• More directly, $$\phi(a) = [a/1] \in \left\{{p,q {~\mathrel{\Big\vert}~}p\in R, q\in A}\right\}$$, which has inverse $$[a/1]$$.

### Spring 2021 #5#algebra/qual/completed

Suppose that $$f(x) \in ({\mathbf{Z}}/n{\mathbf{Z}})[x]$$ is a zero divisor. Show that there is a nonzero $$a\in {\mathbf{Z}}/n{\mathbf{Z}}$$ with $$af(x) = 0$$.

• Write $$f(x) = \sum_{k=0}^n a_k x^k$$, and supposing it’s a zero divisor choose $$g(x) = \sum_{k=0}^m b_k x^k$$ of minimal degree so that $$g\neq 0, b_m\neq 0$$, and $$f(x)g(x) = 0$$.
• The claim is that the top coefficient $$b_m$$ will suffice.
• Write the product: \begin{align*} 0 = f(x)g(x) = (a_0 + \cdots + a_{n-1}x^{n-1} + a_n x^n) (b_0 + \cdots + b_{m-1}x^{m-1} + b_m x^m) .\end{align*}
• Equating coefficients, the coefficient for $$x^{m+n}$$ must be zero, so (importantly) $$a_n b_m = 0$$.
• Since $$a_n b_m=0$$, consider $$a_ng(x)$$. This has degree $$d_1 \leq m-1$$ but satisfies $$a_ng(x)f(x) = a_n(g(x)f(x)) = 0$$, so by minimality $$a_ng(x) = 0$$.
• This forces $$a_n b_0 = \cdots = a_n b_{m-1} = 0$$, so $$a_n$$ annihilates all of the $$b_k$$.
• Now consider the coefficient of $$x^{m+n-1}$$, given by $$a_{n-1}b_m + a_{n}b_{m-1}$$.
• The second term $$a_n b_{m-1}=0$$ since $$a_n$$ annihilates all $$b_k$$, so (importantly) $$a_{n-1} b_m = 0$$.
• Considering now $$a_{n-1}g(x)$$:
• The same argument shows this has degree $$d_2 \leq m-1$$ but $$a_{n-1}g(x)f(x) = 0$$, so $$a_{n-1}g(x) = 0$$.
• So $$a_{n-1}$$ annihilates all $$b_k$$, and allowing this process to continue inductively.
• For good measure, the coefficient of $$x^{m+n-2}$$ is $$a_{n-2}b_m + a_{n-1}b_{m-1} + a_{n}b_{m-2}$$.
• Note that $$a_n, a_{n-1}$$ annihilate all $$b_k$$, so (importantly) $$a_{n-2} b_m=0$$, and so on.
• Thus $$a_k b_m = 0$$ for all $$0\leq k \leq n$$, and by linearity and commutativity, we have \begin{align*} b_m f(x) = b_m \sum_{k=0}^n a_k x^k = \sum_{k=0}^n (b_m a_k) x^k = 0 .\end{align*}

### Fall 2018 #7#algebra/qual/completed

Let $$R$$ be a commutative ring.

• Let $$r \in R$$. Show that the map \begin{align*} r\bullet : R &\to R \\ x &\mapsto r x .\end{align*} is an $$R{\hbox{-}}$$module endomorphism of $$R$$.

• We say that $$r$$ is a zero-divisor if $$r\bullet$$ is not injective. Show that if $$r$$ is a zero-divisor and $$r \neq 0$$, then the kernel and image of $$R$$ each consist of zero-divisors.

• Let $$n \geq 2$$ be an integer. Show: if $$R$$ has exactly $$n$$ zero-divisors, then $${\sharp}R \leq n^2$$ .

• Show that up to isomorphism there are exactly two commutative rings $$R$$ with precisely 2 zero-divisors.

You may use without proof the following fact: every ring of order 4 is isomorphic to exactly one of the following: \begin{align*} \frac{ {\mathbf{Z}}}{ 4{\mathbf{Z}}}, \quad \frac{ \frac{ {\mathbf{Z}}}{ 2{\mathbf{Z}}} [t]}{(t^2 + t + 1)}, \quad \frac{ \frac{ {\mathbf{Z}}}{ 2{\mathbf{Z}}} [t]}{ (t^2 - t)}, \quad \frac{ \frac{ {\mathbf{Z}}}{2{\mathbf{Z}}}[t]}{(t^2 )} .\end{align*}

• Testing module morphisms: $$\phi(sm + n) = s\phi(m) + \phi(n)$$.
• First isomorphism theorem used for sizes: $$R/\ker f \cong \operatorname{im}f$$, so $${\sharp}R = {\sharp}\ker f \cdot {\sharp}\operatorname{im}f$$.
• See 1964 Annals “Properties of rings with a finite number of zero divisors”

• Let $$\phi$$ denote the map in question, it suffices to show that $$\phi$$ is $$R{\hbox{-}}$$linear, i.e. $$\phi(s\mathbf{x} + \mathbf{y}) = s\phi(\mathbf{x}) + \phi(\mathbf{y})$$: \begin{align*} \phi(s\mathbf{x} + \mathbf{y}) &= r(s\mathbf{x} + \mathbf{y}) \\ &= rs\mathbf{x} + r\mathbf{y} \\ &= s(r\mathbf{x}) + (r\mathbf{y}) \\ &= s\phi(\mathbf{x}) + \phi(\mathbf{y}) .\end{align*}

Let $$\phi_r(x) \coloneqq rx$$ be the multiplication map.

• Let $$x\in \ker \phi_r \coloneqq\left\{{x\in R {~\mathrel{\Big\vert}~}rx = 0}\right\}$$.

• Since $$R$$ is commutative $$0 = rx = xr$$, and so $$r\in \ker \phi_x$$, so $$\ker \phi_x \neq 0$$ and $$x$$ is a zero divisor by definition.

• Let $$y\in \operatorname{im}\phi_r \coloneqq\left\{{y \coloneqq rx {~\mathrel{\Big\vert}~}x\in R}\right\}$$, we want to show $$\ker \phi_y$$ is nontrivial by producing some $$z$$ such that $$yz=0$$. Write $$y\coloneqq rx$$ for some $$x\in R$$.

• Since $$r$$ is a zero divisor, we can produce some $$z\neq 0 \in \ker \phi_r$$, so $$rz = 0$$.

• Now using that $$R$$ is commutative, we can compute \begin{align*} yz = (rx)z = (xr)z = x (rz) = x(0) = 0 ,\end{align*} so $$z\in \ker \phi_y$$.

• Let $$Z \coloneqq\left\{{z_i}\right\}_{i=1}^n$$ be the set of $$n$$ zero divisors in $$R$$.

• Let $$\phi_i$$ be the $$n$$ maps $$x \mapsto z_i x$$, and let $$K_i = \ker \phi_i$$ be the corresponding kernels.

• Fix an $$i$$.

• By (b), $$K_i$$ consists of zero divisors, so \begin{align*} {\left\lvert {K_i} \right\rvert} \leq n < \infty \quad \text{for each } i .\end{align*}

• Now consider $$R/K_i \coloneqq\left\{{r + K_i}\right\}$$.

• By the first isomorphism theorem, $$R/K_i \cong \operatorname{im}\phi$$, and by (b) every element in the image is a zero divisor, so \begin{align*} [R: K_i] = {\left\lvert {R/K_i} \right\rvert} = {\left\lvert {\operatorname{im}\phi_i} \right\rvert} \leq n .\end{align*}

• But then \begin{align*} {\left\lvert {R} \right\rvert} = [R:K_i]\cdot {\left\lvert {K_i} \right\rvert} \leq n\cdot n = n^2 .\end{align*}

• By (c), if there are exactly 2 zero divisors then $${\left\lvert {R} \right\rvert} \leq 4$$. Since every element in a finite ring is either a unit or a zero divisor, and $${\left\lvert {R^{\times}} \right\rvert} \geq 2$$ since $$\pm 1$$ are always units, we must have $${\left\lvert {R} \right\rvert} = 4$$.

• Since the characteristic of a ring must divide its size, we have $$\operatorname{ch}R = 2$$ or $$4$$.

• Using the hint, we see that only $${\mathbf{Z}}/(4)$$ has characteristic 4, which has exactly 2 zero divisors given by $$[0]_4$$ and $$[2]_4$$.

• If $$R$$ has characteristic 2, we can check the other 3 possibilities.

• We can write $${\mathbf{Z}}/(2)[t]/(t^2) = \left\{{a + bt {~\mathrel{\Big\vert}~}a,b\in {\mathbf{Z}}/(2)}\right\}$$, and checking the multiplication table we have \begin{align*} \begin{array}{c|cccc} & 0 & 1 & t & 1+t \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & t & 1+t \\ t & 0 & t & \mathbf{0} & t \\ 1 + t & 0 & 1+t & t & 1 \\ \end{array} ,\end{align*}

and so we find that $$t, 0$$ are the zero divisors.

• In $${\mathbf{Z}}/(2)[t]/(t^2 - t)$$, we can check that $$t^2 = t \implies t t^2 = t^2 \implies t(t^2 + 1) = 0 \implies t(t+1) = 0$$, so both $$t$$ and $$t+1$$ are zero divisors, along with zero, so this is not a possibility.

• Similarly, in $${\mathbf{Z}}/(2)[t]/(t^2 + t + 1)$$, we can check the bottom-right corner of the multiplication table to find \begin{align*} \left[\begin{array}{c|cc} & t & 1 +t \\ \hline t & 1+t & 1 \\ t & 1 & t \\ \end{array}\right] ,\end{align*}

and so this ring only has one zero divisor.

• Thus the only possibilities are: \begin{align*} R &\cong {\mathbf{Z}}/(4) \\ R &\cong {\mathbf{Z}}/(2)[t] / (t^2) .\end{align*}

## Zorn’s Lemma

### Fall 2013 #4#algebra/qual/work

Let $$R$$ be a commutative ring with $$1\neq 0$$. Recall that $$x\in R$$ is nilpotent iff $$x^n = 0$$ for some positive integer $$n$$.

• Show that the collection of nilpotent elements in $$R$$ forms an ideal.

• Show that if $$x$$ is nilpotent, then $$x$$ is contained in every prime ideal of $$R$$.

• Suppose $$x\in R$$ is not nilpotent and let $$S = \left\{{x^n {~\mathrel{\Big\vert}~}n\in {\mathbb{N}}}\right\}$$. There is at least on ideal of $$R$$ disjoint from $$S$$, namely $$(0)$$.

By Zorn’s lemma the set of ideals disjoint from $$S$$ has a maximal element with respect to inclusion, say $$I$$. In other words, $$I$$ is disjoint from $$S$$ and if $$J$$ is any ideal disjoint from $$S$$ with $$I\subseteq J \subseteq R$$ then $$J=I$$ or $$J=R$$.

Show that $$I$$ is a prime ideal.

• Deduce from (a) and (b) that the set of nilpotent elements of $$R$$ is the intersection of all prime ideals of $$R$$.

### Fall 2015 #3#algebra/qual/completed

Let $$R$$ be a rng (a ring without 1) which contains an element $$u$$ such that for all $$y\in R$$, there exists an $$x\in R$$ such that $$xu=y$$.

Prove that $$R$$ contains a maximal left ideal.

Hint: imitate the proof (using Zorn’s lemma) in the case where $$R$$ does have a 1.

• Define the map \begin{align*} \phi_u: R &\to R\\ x &\mapsto xu ,\end{align*} which is right-multiplication by $$u$$. By assumption, $$\phi_u$$ is surjective, so the principal ideal $$Ru$$ equals $$R$$.

• Then $$K \coloneqq\ker \phi_u \in \operatorname{Id}(R)$$ is an ideal.

• $$K$$ is proper – otherwise, noting $$Ru=R$$, if $$K=R$$ we have $$Ru = 0$$ and thus $$R=0$$. So suppose $$R\neq 0$$.

• Take a poset $$S \coloneqq\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J \supseteq K, J\neq R}\right\}$$, the set of all ideals containing $$K$$, ordered by subset inclusion. Note that $$K \in S$$, so $$S$$ is nonempty.

• Apply Zorn’s lemma: let $$C: C_1 \subseteq C_2 \subseteq \cdots$$ be a chain (totally ordered sub-poset) in $$S$$. If $$C$$ is the empty chain, $$K$$ is an upper bound. Otherwise, if $$C$$ is nonempty, define $$\widehat{C} \coloneqq\bigcup_{i=1}^\infty C_i$$.

• $$\widehat{C}$$ is an ideal: if $$a,b\in \widehat{C}$$, then $$a\in C_i, b\in C_j$$ for some $$i,j$$. But without loss of generality, using that chains are totally ordered, $$C_i \subseteq C_j$$, so $$a,b\in C_j$$ and thus $$ab\in C_j$$. Similarly for $$x\in \widehat{C}$$, $$x\in C_j$$ for some $$j$$, and thus $$Rx \subseteq C_j$$ since $$C_j$$ is an ideal.
• $$\widehat{C}$$ is in $$S$$: It clearly contains $$K$$, since for example $$K \subseteq C_1 \subseteq \widehat{C}$$.
• That $$\widehat{C} \neq R$$: an ideal equals $$R$$ iff it contains a unit. But if $$r\in \widehat{C}$$ is a unit, $$r\in C_j$$ for some $$j$$ is a unit, making $$C_j=R$$. $$\contradiction$$
• So by Zorn’s lemma, $$\widehat{C}$$ contains a maximal ideal (incidentally containing $$K$$).

### Spring 2015 #7#algebra/qual/completed

Let $$R$$ be a commutative ring, and $$S\subset R$$ be a nonempty subset that does not contain 0 such that for all $$x, y\in S$$ we have $$xy\in S$$. Let $${\mathcal{I}}$$ be the set of all ideals $$I{~\trianglelefteq~}R$$ such that $$I\cap S = \emptyset$$.

Show that for every ideal $$I\in {\mathcal{I}}$$, there is an ideal $$J\in {\mathcal{I}}$$ such that $$I\subset J$$ and $$J$$ is not properly contained in any other ideal in $${\mathcal{I}}$$.

Prove that every such ideal $$J$$ is prime.

• Restating, take the poset $$S\coloneqq\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J \cap S = \emptyset, I\neq R, I \subseteq J}\right\}$$ ordered by inclusion. Note that $$S$$ is nonempty since it contains $$I$$. It suffices to produce a maximal element of $$S$$.
• Applying Zorn’s lemma, let $$C: C_1 \subseteq C_2 \subseteq \cdots$$ be a chain and define $$\widehat{C} \coloneqq\cup C_i$$.
• By standard arguments, $$\widehat{C} \in \operatorname{Id}(R)$$ and $$\widehat{C} \supseteq I$$, and it suffices to show $$\widehat{C} \cap S = \emptyset$$ and $$\widehat{C}\neq R$$.
• $$\widehat{C} \cap S = \emptyset$$:
• By contradiction, if $$x\in \widehat{C} \cap S$$ then $$x\in C_j$$ for some $$j$$, and $$x\in S$$. But then $$x \in C_j \cap S = \emptyset$$.
• $$\widehat{C} \neq R$$:
• By contradiction, if so then $$1 \in \widehat{C} \implies 1 \in C_j$$ for some $$j$$, forcing $$C_j = R$$.
• So Zorn’s lemma applies and we obtain some ideal $${\mathfrak{p}}$$, which we now claim is prime.
• Let $$ab\in {\mathfrak{p}}$$, we want to show $$a\in {\mathfrak{p}}$$ or $$b\in{\mathfrak{p}}$$.
• Suppose not, then neither $$a,b\in {\mathfrak{p}}$$. By maximality, $${\mathfrak{p}}+ Ra = R$$, and so $${\mathfrak{p}}+ Ra$$ intersects $$S$$. Similarly $${\mathfrak{p}}+ Rb = R$$ so $${\mathfrak{p}}+ Rb$$ intersects $$S$$.
• Produce elements $$x\coloneqq p_1 + r_1a, y\coloneqq p_2 + r_2b\in S$$, then since $$S$$ is multiplicatively closed, \begin{align*} xy&\coloneqq(p_1 + r_1 a)(p_2 + r_2b)\in S \\ &\implies p_1 p_2 + p_1r_2 b + p_2 r_1 a + r_1 r_2 ab \in S \\ &\implies xy\in {\mathfrak{p}}+ {\mathfrak{p}}Rb + {\mathfrak{p}}Ra + R{\mathfrak{p}}&& \text{since } p_i, ab\in {\mathfrak{p}}\\ &\implies xy \in ({\mathfrak{p}}+ Rb + Ra + R){\mathfrak{p}}\subseteq {\mathfrak{p}} .\end{align*} But then $$xy\in S \cap{\mathfrak{p}}$$, a contradiction.

### Spring 2013 #1#algebra/qual/completed

Let $$R$$ be a commutative ring.

• Define a maximal ideal and prove that $$R$$ has a maximal ideal.

• Show than an element $$r\in R$$ is not invertible $$\iff r$$ is contained in a maximal ideal.

• Let $$M$$ be an $$R{\hbox{-}}$$module, and recall that for $$0\neq \mu \in M$$, the annihilator of $$\mu$$ is the set \begin{align*} \operatorname{Ann}(\mu) = \left\{{r\in R {~\mathrel{\Big\vert}~}r\mu = 0}\right\} .\end{align*} Suppose that $$I$$ is an ideal in $$R$$ which is maximal with respect to the property that there exists an element $$\mu \in M$$ such that $$I = \operatorname{Ann}(\mu)$$ for some $$\mu \in M$$. In other words, $$I = \operatorname{Ann}(\mu)$$ but there does not exist $$\nu\in M$$ with $$J = \operatorname{Ann}(\nu) \subsetneq R$$ such that $$I\subsetneq J$$.

Prove that $$I$$ is a prime ideal.

• Maximal: a proper ideal $$I{~\trianglelefteq~}R$$, so $$I\neq R$$, such that if $$J\supseteq I$$ is any other ideal, $$J = R$$.
• Existence of a maximal ideal: use that $$0\in \operatorname{Id}(R)$$ always, so $$S\coloneqq\left\{{I\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}I\neq R}\right\}$$ is a nonempty poset under subset inclusion. Applying the usual Zorn’s lemma argument produces a maximal element.

$$\impliedby$$: By contrapositive: if $$r\in R$$ is a unit and $${\mathfrak{m}}$$ is maximal, then $$r\in {\mathfrak{m}}\implies {\mathfrak{m}}= R$$, contradicting that $${\mathfrak{m}}$$ is proper.

$$\implies$$:

• Suppose $$a$$ is not a unit, we’ll produce a maximal ideal containing it.
• Let $$I\coloneqq Ra$$ be the principal ideal generated by $$a$$, then $$Ra \neq R$$ since $$a$$ is not a unit.
• Take a poset $$S \coloneqq\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J\supseteq Ra, J\neq R}\right\}$$ ordered by set inclusion.
• Let $$C_*$$ be a chain in $$S$$, set $$\widehat{C} \coloneqq\cup C_i$$. Then $$\widehat{C} \in S$$:

• $$\widehat{C} \neq R$$, since if so it contains a unit, forcing some $$C_i$$ to contain a unit and thus equal $$R$$.
• $$\widehat{C} \supseteq Ra$$, since e.g. $$\widehat{C} \supseteq C_1 \supseteq Ra$$.
• $$\widehat{C}$$ is an ideal since $$xy\in \widehat{C} \implies x\in C_i, y\in C_j$$ and $$C_i \subseteq C_j$$ without loss of generality, so $$xy\in C_j \subseteq \widehat{C}$$.
• Then $$Ra \subseteq \widehat{C}$$, some maximal ideal.

• Write $$I \coloneqq\operatorname{Ann}(u)$$ for some $$u$$, and toward a contradiction suppose $$ab\in I$$ but $$a,b\not\in I$$.
• Then $$abu=0$$ but $$au\neq 0, bu\neq 0$$.
• Since $$abu=0$$, we have $$a\in \operatorname{Ann}(bu)$$. Note that $$\operatorname{Ann}(bu) \supseteq\operatorname{Ann}(u)$$, since $$su = 0\implies bsu=sbu=0$$.
• We can’t have $$\operatorname{Ann}(bu) = R$$, since if $$sbu=0$$ for all $$s\in R$$, so we could take $$s=1$$ to get $$bu=0$$ and $$b\in \operatorname{Ann}(u)$$.
• By maximality, this forces $$\operatorname{Ann}(u) = \operatorname{Ann}(bu)$$, so $$sbu = 0 \implies su=0$$ for any $$s\in R$$.
• Now take $$s=a$$, and since $$abu=0$$ we get $$au=0$$ and $$a\in \operatorname{Ann}(u)$$. $$\contradiction$$

### Fall 2019 #6#algebra/qual/completed

Let $$R$$ be a commutative ring with multiplicative identity. Assume Zorn’s Lemma.

• Show that \begin{align*} N = \{r \in R \mathrel{\Big|}r^n = 0 \text{ for some } n > 0\} \end{align*} is an ideal which is contained in any prime ideal.

• Let $$r$$ be an element of $$R$$ not in $$N$$. Let $$S$$ be the collection of all proper ideals of $$R$$ not containing any positive power of $$r$$. Use Zorn’s Lemma to prove that there is a prime ideal in $$S$$.

• Suppose that $$R$$ has exactly one prime ideal $$P$$ . Prove that every element $$r$$ of $$R$$ is either nilpotent or a unit.

• Prime ideal: $$\mathfrak{p}$$ is prime iff $$ab \in \mathfrak{p} \implies a\in \mathfrak{p}$$ or $$b\in \mathfrak{p}$$.

• Silly fact: 0 is in every ideal!

• Zorn’s Lemma: Given a poset, if every chain has an upper bound, then there is a maximal element. (Chain: totally ordered subset.)

• Corollary: If $$S\subset R$$ is multiplicatively closed with $$0\not\in S$$ then $$\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big\vert}~}J\cap S = \emptyset}\right\}$$ has a maximal element.

\todo[inline]{Prove this}

• Theorem: If $$R$$ is commutative, maximal $$\implies$$ prime for ideals.

\todo[inline]{Prove this}

• Theorem: Non-units are contained in a maximal ideal. (See HW?)

• Let $$\mathfrak{p}$$ be prime and $$x\in N$$.
• Then $$x^k = 0 \in \mathfrak{p}$$ for some $$k$$, and thus $$x^k = x x^{k-1} \in \mathfrak p$$.
• Since $$\mathfrak p$$ is prime, inductively we obtain $$x\in\mathfrak p$$.

• Let $$S = \left\{{r^k \mathrel{\Big|}k\in {\mathbb{N}}}\right\}$$ be the set of positive powers of $$r$$.

• Then $$S^2 \subseteq S$$, since $$r^{k_1}r^{k_2} = r^{k_1+k_2}$$ is also a positive power of $$r$$, and $$0\not\in S$$ since $$r\neq 0$$ and $$r\not\in N$$.

• By the corollary, $$\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big\vert}~}I\cap S = \emptyset}\right\}$$ has a maximal element $$\mathfrak{p}$$.

• Since $$R$$ is commutative, $$\mathfrak{p}$$ is prime.

• Suppose $$R$$ has a unique prime ideal $$\mathfrak{p}$$.

• Suppose $$r\in R$$ is not a unit, and toward a contradiction, suppose that $$r$$ is also not nilpotent.

• Since $$r$$ is not a unit, $$r$$ is contained in some maximal (and thus prime) ideal, and thus $$r \in \mathfrak{p}$$.

• Since $$r\not\in N$$, by (b) there is a maximal ideal $$\mathfrak{m}$$ that avoids all positive powers of $$r$$. Since $$\mathfrak{m}$$ is prime, we must have $$\mathfrak{m} = \mathfrak{p}$$. But then $$r\not\in \mathfrak{p}$$, a contradiction.

## Noetherian Rings

### Fall 2015 #4#algebra/qual/completed

Let $$R$$ be a PID and $$(a_1) < (a_2) < \cdots$$ be an ascending chain of ideals in $$R$$. Prove that for some $$n$$, we have $$(a_j) = (a_n)$$ for all $$j\geq n$$.

• Let $$I\coloneqq\cup Ra_i$$ which is an ideal in a PID and thus $$I = Rb$$ for some $$b$$.
• Using that $$b\in I$$, which is a union, we have $$Rb\in Ra_m$$ for some $$m$$.
• Thus $$I = R_b \subseteq Ra_m$$, and $$Ra_m \subseteq I$$ by definition of $$I$$, so $$Rb = Ra_m$$.
• In particular, since $$Ra_{m} \subseteq Ra_{m+1}$$ by assumption, and $$Ra_{m+1} \subseteq Rb \subseteq Ra_m$$ since $$Rb = I$$, we have $$Ra_m = Ra_{m+1}$$. So inductively, the chain stabilizes at $$m$$.

### Spring 2021 #6#algebra/qual/work

• Carefully state the definition of Noetherian for a commutative ring $$R$$.

• Let $$R$$ be a subset of $${\mathbf{Z}}[x]$$ consisting of all polynomials \begin{align*} f(x) = a_ 0 + a_1 x + a_2 x^2 + \cdots + a_nx^n \end{align*} such that $$a_k$$ is even for $$1\leq k \leq n$$. Show that $$R$$ is a subring of $${\mathbf{Z}}[x]$$.

• Show that $$R$$ is not Noetherian.

Hint: consider the ideal generated by $$\left\{{ 2x^k {~\mathrel{\Big\vert}~}1\leq k \in {\mathbf{Z}}}\right\}$$.

• A ring is Noetherian iff $$R$$ satisfies the ascending chain condition: every chain of ideals $$A_1 \subseteq A_2 \subseteq \cdots$$ eventually stabilizes, so $$A_m \subseteq A_{m+1} = A_{m+2} = \cdots$$.

• That $$R$$ is a subring of $${\mathbf{Z}}[x]$$:

• $$(R, +)$$ is an abelian subgroup: note that $$f(x) + g(x) = \sum a_k x^k + \sum b_k x^k = \sum (a_k + b_k) x^k$$, so if $$a_k, b_k$$ are even then $$a_k + b_k$$ is even. It’s closed under inverses, since $$a_k$$ is even iff $$-a_k$$ is even, and contains zero.
• $$(R, \cdot)$$ is a submonoid: $$f(x) g(x) = \sum_{n=1}^N \qty{ \sum_{k=1}^n a_k b_{n-k}} x^k$$ where without loss of generality, $$\deg f = \deg g = n$$ by setting coefficients to zero. Then sums and products of even integers are even, so $$fg \in R$$.
• That $$R$$ is not Noetherian: it suffices to show that $$R$$ contains an ideal that is not finitely generated.

• The claim is that setting $$S \coloneqq\left\{{2x^k}\right\}_{k\in {\mathbf{Z}}_{\geq 1}}$$ and taking \begin{align*} I \coloneqq\left\langle{S}\right\rangle = \sum_{k\in {\mathbf{Z}}_{\geq 1}} R\cdot 2x^k \coloneqq\left\{{ \sum_{i=1}^m r_k(x) 2x^k {~\mathrel{\Big\vert}~}r_k(x) \in 2{\mathbf{Z}}[x], m\in {\mathbf{Z}}_{\geq 0}}\right\} \end{align*} yields an ideal that is not finitely generated.

• Suppose toward a contradiction that $$\left\{{g_1, g_2, \cdots, g_M}\right\}$$ were a finite generating set, where each $$g_i \in I$$.

\todo[inline]{???}


## Simple Rings

### Fall 2017 #5#algebra/qual/completed

A ring $$R$$ is called simple if its only two-sided ideals are $$0$$ and $$R$$.

• Suppose $$R$$ is a commutative ring with 1. Prove $$R$$ is simple if and only if $$R$$ is a field.

• Let $$k$$ be a field. Show the ring $$M_n (k)$$, $$n \times n$$ matrices with entries in $$k$$, is a simple ring.

• Nonzero proper ideals contain at least one nonzero element.
• $$I=R$$ when $$1\in I$$.
• Effects of special matrices: let $$A_{ij}$$ be a matrix with only a 1 in the $$ij$$ position.
• Left-multiplying $$A_{ij}M$$ moves row $$j$$ to row $$i$$ and zeros out the rest of $$M$$.
• Right-multiplying $$MA_{ij}$$ moves column $$i$$ to column $$j$$ and zeros out the rest.
• So $$A_{ij} M A_{kl}$$ moves entry $$j, k$$ to $$i, l$$ and zeros out the rest.

$$\implies$$:

• Suppose $$\operatorname{Id}(R) = \left\{{\left\langle{0}\right\rangle, \left\langle{1}\right\rangle}\right\}$$. Then for any nonzero $$r\in R$$, the ideal $$\left\langle{r}\right\rangle = \left\langle{1}\right\rangle$$ is the entire ring.
• In particular, $$1\in \left\langle{r}\right\rangle$$, so we can write $$a = tr$$ for some $$t\in R$$.
• But then $$r\in R^{\times}$$ with $$t\coloneqq r^{-1}$$.

$$\impliedby$$:

• Suppose $$R$$ is a field and $$I\in \operatorname{Id}(R)$$ is an ideal.
• If $$I \neq \left\langle{0}\right\rangle$$ is proper and nontrivial, then $$I$$ contains at least one nonzero element $$a\in I$$.
• Since $$R$$ is a field, $$a^{-1}\in R$$, and $$aa^{-1}= 1\in I$$ forces $$I = \left\langle{1}\right\rangle$$.
• Let $$J{~\trianglelefteq~}R$$ be a nonzero two-sided ideal, noting that $$R$$ is noncommutative – the claim is that $$J$$ contains $$I_n$$, the $$n\times n$$ identity matrix, and thus $$J = R$$.
• Pick a nonzero element $$M\in I$$, then $$M$$ has a nonzero entry $$m{ij}$$.
• Let $$A_{ij}$$ be the matrix with a 1 in the $$i,j$$ position and zeros elsewhere.
• Left-multiplying $$A_{ij}M$$ moves row $$j$$ to row $$i$$ and zeros out the rest of $$M$$.
• Right-multiplying $$MA_{ij}$$ moves column $$i$$ to column $$j$$ and zeros out the rest.
• So $$A_{ij} M A_{kl}$$ moves entry $$j, k$$ to $$i, l$$ and zeros out the rest.
• So define $$B \coloneqq A_{i, i}MA_{j, i}$$, which movies $$m_{ij}$$ to the $$i,i$$ position on the diagonal and has zeros elsewhere.
• Then $$m_{ij}^{-1}{\varepsilon}_{ii} \coloneqq A_{ii}$$ is a matrix with $$1$$ in the $$i, i$$ spot for any $$i$$. Since $$I$$ is an ideal, we have $${\varepsilon}_{ii}\in I$$ for every $$i$$.
• We can write the identity $$I_n$$ as $$\sum_{i=1}^n {\varepsilon}_{ii}$$, so $$I_n \in I$$ and $$I=R$$.

### Spring 2016 #8#algebra/qual/completed

Let $$R$$ be a simple rng (a nonzero ring which is not assume to have a 1, whose only two-sided ideals are $$(0)$$ and $$R$$) satisfying the following two conditions:

• $$R$$ has no zero divisors, and
• If $$x\in R$$ with $$x\neq 0$$ then $$2x\neq 0$$, where $$2x\coloneqq x+x$$.

Prove the following:

• For each $$x\in R$$ there is one and only one element $$y\in R$$ such that $$x = 2y$$.

• Suppose $$x,y\in R$$ such that $$x\neq 0$$ and $$2(xy) = x$$, then $$yz = zy$$ for all $$z\in R$$.

You can get partial credit for (b) by showing it in the case $$R$$ has a 1.

A general opinion is that this is not a great qual problem! Possibly worth skipping.

• $$R$$ has no left zero divisors iff $$R$$ has the left cancellation property: $$xa=xb \implies a=b$$.
• $$R$$ has no right zero divisors iff $$R$$ has the right cancellation property: $$ax=bx \implies a=b$$.

Note: solutions borrowed from folks on Math twitter!

• Existence: the claim is that $$2R \coloneqq\left\{{2y {~\mathrel{\Big\vert}~}y\in R}\right\}$$ is a nontrivial two-sided ideal of $$R$$, forcing $$2R = R$$ by simpleness.
• That $$2R \neq 0$$ follows from condition (1): Provided $$y\neq 0$$, we have $$2y\neq 0$$, and so if $$R\neq 0$$ then there exists some nonzero $$a\in R$$, in which case $$2a\neq 0$$ and $$2a\in 2R$$.
• That $$2R$$ is a right ideal: clear, since $$(2y)\cdot r = 2(yr)\in 2R$$.
• That $$2R$$ is a left ideal: use that multiplication is distributive: \begin{align*} r\cdot 2y \coloneqq r(y+y) = ry + ry \coloneqq 2(ry) \in 2R .\end{align*}
• So $$2R = R$$ by simpleness.
• Uniqueness:
• Use the contrapositive of condition (1), so that $$2x = 0 \implies x=0$$.
• Suppose toward a contradiction that $$x=2y_1 = 2y_2$$, then \begin{align*} 0 = x-x = 2y_1 - 2y_2 = 2(y_1 - y_2) \implies y_1 - y_2 = 0 \implies y_1 = y_2 .\end{align*}

• First we’ll show $$z=2(yz)$$: \begin{align*} xy + xy &= x \\ \implies xy + xy - x &= 0 \\ \implies xyz + xyz - xz &= 0 \\ \implies x(yz + yz - z) &= 0 \\ \implies yz + yz - z &= 0 && \text{since } x\neq 0 \text{ and no zero divisors }\\ \implies 2(yz) &= z .\end{align*}

• Now we’ll show $$z=2(zy)$$: \begin{align*} yz + yz &= z \\ \implies zyz + zyz &= zz \\ \implies zyz + zyz - zz &= 0 \\ \implies (zy + zy - z)z &= 0\\ \implies z=0 \text{ or } zy+zy-z &= 0 && \text{ no zero divisors } .\end{align*}

• Then if $$z=0$$, we have $$yz = 0 = zy$$ and we’re done.

• Otherwise, $$2(zy) = z$$, and thus \begin{align*} 2(zy) = z = 2(yz) \implies 2(zy - yz) = 0 \implies zy-yz = 0 ,\end{align*} so $$zy=yz$$.

• If $$1\in R$$, \begin{align*} 2xy &= x \\ \implies 2xy-x &= 0 \\ \implies x(2y-1) &= 0 \\ \implies 2y-1 &= 0 && x\neq 0 \text{ and no zero divisors}\\ \implies 2y &= 1 .\end{align*}
• Now use \begin{align*} 1\cdot z &= z\cdot 1 \\ \implies (2y)z &= z(2y) \\ \implies (y+y)z &= z(y+y) \\ \implies yz+yz &= zy+zy \\ \implies 2(yz) &= 2(zy) \\ \implies 2(yz-zy) &= 0 \\ \implies yz-zy &= 0 \\ ,\end{align*} using condition (2).

## Unsorted

### Fall 2019 #3#algebra/qual/completed

Let $$R$$ be a ring with the property that for every $$a \in R, a^2 = a$$.

• Prove that $$R$$ has characteristic 2.

• Prove that $$R$$ is commutative.

• Just fiddle with direct computations.
• Context hint: that we should be considering things like $$x^2$$ and $$a+b$$.

\begin{align*} 2a = (2a)^2 = 4a^2 = 4a \implies 2a = 0 .\end{align*} Note that this implies $$x = -x$$ for all $$x\in R$$.

\begin{align*} a+b = (a+b)^2 &= a^2 + ab + ba + b^2 = a + ab + ba + b \\ &\implies ab + ba = 0 \\ &\implies ab = -ba \\ &\implies ab = ba \quad\text{by (a)} .\end{align*}

### Spring 2018 #5#algebra/qual/completed

Let \begin{align*} M=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right) \quad \text{and} \quad N=\left(\begin{array}{cc}{x} & {u} \\ {-y} & {-v}\end{array}\right) \end{align*}

over a commutative ring $$R$$, where $$b$$ and $$x$$ are units of $$R$$. Prove that \begin{align*} M N=\left(\begin{array}{ll}{0} & {0} \\ {0} & {*}\end{array}\right) \implies MN = 0 .\end{align*}

• Multiply everything out to get \begin{align*} { \begin{bmatrix} {ax-by} & {au-bv} \\ {cx-dy} & {cu-dv} \end{bmatrix} } ,\end{align*} so it suffices to show $$cu=dv$$ given \begin{align*} ax &= by \\ cx &= dy \\ au &= bv .\end{align*}

• Writing $$cu$$:

• Use that $$b\in R^{\times}$$, left-multiply (1) by $$b^{-1}$$ to get $$b^{-1}a x = y$$
• Substitute $$y$$ into (2) to get $$cx = d(b^{-1}a x)$$.
• Since $$x\in R^{\times}$$, right-multiply by $$x^{-1}$$ to get $$c = db^{-1}a$$ and thus $$cu = db^{-1}a u$$.
• Summary: \begin{align*} ax = by &\implies b^{-1}ax = y \\ &\implies cx = dy = d(b^{-1}a x) \\ &\implies c = db^{-1}a \\ &\implies cu = db^{-1}au .\end{align*}
• Writing $$dv$$:

• Left-multiply (3) by $$b^{-1}$$ to get $$b^{-1}au = v$$.
• Left-multiply by $$d$$ to get $$db^{-1}au = dv$$
• Summary: \begin{align*} au = bv &\implies b^{-1}a u = v \\ &\implies db^{-1}au = dv .\end{align*}
• So \begin{align*} cu = db^{-1}a u = dv .\end{align*}

### Spring 2014 #6#algebra/qual/work

$$R$$ be a commutative ring with identity and let $$n$$ be a positive integer.

• Prove that every surjective $$R{\hbox{-}}$$linear endomorphism $$T: R^n \to R^n$$ is injective.

• Show that an injective $$R{\hbox{-}}$$linear endomorphism of $$R^n$$ need not be surjective.

#2 #algebra/qual/completed #6 #4 #5 #algebra/qual/stuck #3 #8 #algebra/qual/work #7 #1