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 \(abq \in I \implies r\in I\).
 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 righthand side product must be empty.
 So \(m=n\) and all \(p_i, q_i\) are associate, QED.

Euclidean implies PID:

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 nonzero, nonunit \(r \in R\) 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 \(\mkern 1.5mu\overline{\mkern1.5mu\alpha\mkern1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern1.5mu\beta\mkern1.5mu}\mkern 1.5mu\) be their conjugates.
 Then \begin{align*} 0 = \alpha \beta = \alpha\mkern 1.5mu\overline{\mkern1.5mu\alpha \mkern1.5mu}\mkern 1.5mu\beta\mkern 1.5mu\overline{\mkern1.5mu\beta \mkern1.5mu}\mkern 1.5mu= (a^25b^2)(c^25d^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 \(N(\alpha) = \alpha \mkern 1.5mu\overline{\mkern1.5mu\alpha \mkern1.5mu}\mkern 1.5mu= (a+b\sqrt{5})(ab\sqrt{5}) = a^2 + 5b^2 = 1\) 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 \(p(x) = x^22\).
 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 \(p(x) \coloneqq x^22 \in R[x]\) 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/work
Let \(R\) be an algebra over \(\mathbb{C}\) which is finitedimensional 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 \(\mathbb{C}{\hbox{}}\)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 \(\operatorname{dim}_{C} R=2\). 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.
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}^{n1} (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 \(1x\) is not a unit, then \(\left\langle{1x}\right\rangle\) is a nontrivial proper ideal, which by (a) is contained in some maximal ideal \({\mathfrak{m}}\). But then \(x\in {\mathfrak{m}}\) and \(1x \in {\mathfrak{m}}\implies x + (1x) = 1 \in {\mathfrak{m}}\), a contradiction.

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

Then \begin{align*} (1x)x &= x  x^2 = x  x = 0 \\ &\implies u (1x)x = x = 0 \\ &\implies x=0 .\end{align*}
Spring 2018 #8 #algebra/qual/completed
Let \(R = C[0, 1]\) be the ring of continuous realvalued 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 \({\left\lvert {xa} \right\rvert}<{\varepsilon}\implies {\left\lvert {f(x)  f(a)} \right\rvert} < c/2\). 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 nonnegative, 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 welldefined: 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_{n1}x^{n1} + a_n x^n) (b_0 + \cdots + b_{m1}x^{m1} + 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 \(d_1 \leq m1\) but satisfies \(a_ng(x)f(x) = a_n(g(x)f(x)) = 0\), so by minimality \(a_ng(x) = 0\).
 This forces \(a_n b_0 = \cdots = a_n b_{m1} = 0\), so \(a_n\) annihilates all of the \(b_k\).

Now consider the coefficient of \(x^{m+n1}\), given by \(a_{n1}b_m + a_{n}b_{m1}\).
 The second term \(a_n b_{m1}=0\) since \(a_n\) annihilates all \(b_k\), so (importantly) \(a_{n1} b_m = 0\).

Considering now \(a_{n1}g(x)\):
 The same argument shows this has degree \(d_2 \leq m1\) but \(a_{n1}g(x)f(x) = 0\), so \(a_{n1}g(x) = 0\).
 So \(a_{n1}\) annihilates all \(b_k\), and allowing this process to continue inductively.

For good measure, the coefficient of \(x^{m+n2}\) is \(a_{n2}b_m + a_{n1}b_{m1} + a_{n}b_{m2}\).
 Note that \(a_n, a_{n1}\) annihilate all \(b_k\), so (importantly) \(a_{n2} b_m=0\), 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 zerodivisor if \(r\bullet\) is not injective. Show that if \(r\) is a zerodivisor and \(r \neq 0\), then the kernel and image of \(R\) each consist of zerodivisors.

Let \(n \geq 2\) be an integer. Show: if \(R\) has exactly \(n\) zerodivisors, then \({\sharp}R \leq n^2\) .

Show that up to isomorphism there are exactly two commutative rings \(R\) with precisely 2 zerodivisors.
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}{ccccc} & 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 bottomright corner of the multiplication table to find \begin{align*} \left[\begin{array}{ccc} & 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 rightmultiplication 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 subposet) 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: Nonunits 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 \(x^k = x x^{k1} \in \mathfrak p\).
 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: \(f(x) g(x) = \sum_{n=1}^N \qty{ \sum_{k=1}^n a_k b_{nk}} x^k\) 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 twosided 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.
 Leftmultiplying \(A_{ij}M\) moves row \(j\) to row \(i\) and zeros out the rest of \(M\).
 Rightmultiplying \(MA_{ij}\) 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 twosided 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.
 Leftmultiplying \(A_{ij}M\) moves row \(j\) to row \(i\) and zeros out the rest of \(M\).
 Rightmultiplying \(MA_{ij}\) 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 twosided 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 twosided 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 = xx = 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+zyz &= 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 zyyz = 0 ,\end{align*} so \(zy=yz\).
 If \(1\in R\), \begin{align*} 2xy &= x \\ \implies 2xyx &= 0 \\ \implies x(2y1) &= 0 \\ \implies 2y1 &= 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(yzzy) &= 0 \\ \implies yzzy &= 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} {axby} & {aubv} \\ {cxdy} & {cudv} \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 \(b\in R^{\times}\), leftmultiply (1) by \(b^{1}\) to get \(b^{1}a x = y\)
 Substitute \(y\) into (2) to get \(cx = d(b^{1}a x)\).
 Since \(x\in R^{\times}\), rightmultiply 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 \(dv\):
 Leftmultiply (3) by \(b^{1}\) to get \(b^{1}au = v\).
 Leftmultiply 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.