A Writeup of the UCSD Spring 2024 Algebra Qualifying Exam

Introduction

Back in my third year as an undergrad at UC San Diego, I had the pleasure of taking Math 200ABC with Alireza Salehi Golsefidy (200AB) and Alina Bucur (200C).

At the end of the year, perhaps against my better judgement, I decided to take the Algebra qualifying exam, which I luckily got a Ph.D. Area Pass on!

Look Mom, I passed!

Now that it’s been a while since I took that exam, I thought it would be fun for posterity if I wrote up some solutions to the problems that came up on that exam and fixed the mistakes I made.

If you’re a future UCSD student that is studying for the Algebra qualifying exam, good luck! And I hope this helps!

Be warned: I make no claims that these solutions are 100% correct, nor that these are the most elegant. Maybe the best use of this writeup is as a source of hints if you’re attempting these problems and you’re stuck.

A link to the PDF version of the exam that I’m referring to can be found here.

Problem 1

Suppose $p$ and $q$ are two distinct primes and $G$ is a group of order $p^2q$. Prove that $G$ is solvable.

Solution

Since $p$ and $q$ are distinct, we consider two cases: $p < q$ and $p > q$. Throughout this solution, we will use $n_p$ and $n_q$ to denote the number of Sylow-$p$ and Sylow-$q$ subgroups, respectively.

Suppose $p > q$. By the Third Sylow Theorem, we obtain $n_p \equiv 1 \pmod p$ and $n_p \mid q$. Since the only factors of $q$ are $1$ and $q$, we immediately obtain $n_p = 1$, as $q \not\equiv 1 \pmod p$, as $q < p$ by assumption. Thus, $G$ has a (unique) normal Sylow-$p$ subgroup. Call this subgroup $P$, so that we have the normal series $1 \trianglelefteq P \trianglelefteq G$. Then, note that $\frac{\lvert G\rvert}{\lvert P\rvert} = q$, so $G / P \isom C_q$ is abelian. And, recall that any group of order $p^2$ is abelian, so $P$ is also abelian. Thus, the factor groups are abelian, so $G$ is solvable.

On the other hand, suppose $q > p$. Then, $n_q \equiv 1 \pmod q$ and $n_q \mid p^2$. The only factors of $p^2$ are $1, p, p^2$. By similar logic as above, since $q > p$, it cannot be the case that $n_q = p$. That is, $n_q$ is either $1$ or $p^2$. If $n_q = 1$, then we are done by a similar argument as above: there is a (unique) normal Sylow-$q$ subgroup $Q$, and we obtain the normal series $1 \trianglelefteq Q \trianglelefteq G$, with factor groups $Q$ (which is cyclic, hence abelian) and a group of order $p^2$ (which is again abelian).

Now, suppose $n_q = p^2$. Let $\Omega$ be the set of elements not order $q$. Then, we calculate that

\begin{equation*} |\Omega| = p^2q - p^2(q - 1) = p^2 \end{equation*}

noting here that each Sylow-$q$ subgroup here must have trivial intersection (they are cyclic of order $q$). By the First Sylow theorem, there must exist a Sylow-$p$ subgroup $P$ with order $p^2$. By Lagrange’s theorem, we obtain $P \subseteq \Omega$. But, since they have the same size, we know that $P = \Omega$ must be the unique normal Sylow-$p$ subgroup of $G$. Repeating the argument from the $p > q$ case, we obtain that $G$ must be solvable.

Thus, any group of order $p^2q$ is solvable.

\[\tag*{$\blacksquare$}\]

Remark

In homework for 200A, we proved a more general statement that a group of order $p^nq$ is solvable (and mentioned that it is true that a group of order $p^nq^m$ is solvable), but I don’t think that proof would have been feasible to replicate during the exam. I also wrote a small argument on the exam about why a group of order $p^2$ is solvable: there’s a standard exercise that if $G/Z(G)$ is cyclic, then $G$ is abelian and that all $p$-groups have non-trivial centers. Hence, any group $P$ of order $p^2$ has $\frac{\lvert P\rvert}{\lvert Z(P)\rvert} \in \{1, p\}$, so $P/Z(P)$ is cyclic and thus $P$ is abelian.

Problem 2

Suppose $G$ is a finite group. Let $\Syl_p(G)$ be the set of Sylow p-subgroups of $G$ and $s_p$ the number of elements in $\Syl_p(G)$.

Part A

Suppose $P, Q \in \Syl_p(G)$ are distinct. Prove that \begin{equation*} P \cap N_G(Q) = P \cap Q \end{equation*}

Solution

Clearly, $P \cap Q \subseteq P \cap N_G(Q)$, as $Q \subseteq N_G(Q)$. On the other hand, $P \cap N_G(Q)$ is a $p$-group by Lagrange’s Theorem (as it is a subgroup of $P$), so $P \cap N_G(Q) \subseteq Q$ is contained in a Sylow-$p$ subgroup of $N_G(Q)$. But, $Q$ is the unique Sylow-$p$ subgroup of $N_G(Q)$, hence $P \cap N_G(Q) \subseteq Q$, so that $P \cap N_G(Q) \subseteq P \cap Q$.

Thus, we obtain the desired equality. \begin{equation*} P \cap N_G(Q) = P \cap Q \end{equation*}

$\tag*{$\blacksquare$}$

Part B

Suppose $P \in \Syl_p(G)$ and consider the action of $P$ on $\Syl_p(G)$ by conjugation. Prove that the $P$-orbit of $Q \in \Syl_p(G)$ has $[P : P \cap Q]$ many elements.

Solution

We see that \begin{align*} g \in P \cap \Stab(Q) &\iff g \in P,\, gQg^{-1} = Q \\ &\iff g \in P \cap N_G(Q) = P \cap Q \end{align*} where the last equality follows from Part A. Hence, by the Orbit-Stabilizer Theorem, \begin{equation*} |\Orb(Q)| = \frac{|P|}{|\Stab(Q)|} = [P : P \cap Q] \end{equation*} as desired.

\[\tag*{$\blacksquare$}\]

Part C

Suppose $p^e \mid (s_p - 1)$ and $p^{e+1} \nmid (s_p - 1)$. Prove that there are distinct $P, Q \in \Syl_p(G)$ such that $[P : P \cap Q] \leq p^{e}$.

Solution

Fix $P \in G$. By Part B, the only fixed point of the conjugation action of $P$ on $\Syl_p(G)$ is $P$ (as $Q$ being fixed implies $|P \cap Q| = |P|$). Suppose, on the contrary, that for all $Q \neq P$, $[P : P \cap Q] > p^e$, i.e. $p^{e+1} \mid [P : P \cap Q]$. That is, $p^{e+1}$ divides $|\Orb(Q)|$ for $Q \neq P$. Then, \begin{align*} s_p &= |\Fix(G)| + \sum_{\OO_i \in \{\OO \in \mathrm{Orbits} \, : \, |\OO| > 1\}} |\OO_i| \end{align*} In particular, by the earlier discussion, $|\Fix(G)| = 1$. Since $p^{e+1} \mid [P : P \cap Q_i]$ for all $i$ and $[P : P \cap Q_i]$ is the size of orbit $\OO_i$ (where $Q_i \in \OO_i$), we obtain $p^{e + 1} \mid s_p - 1$, a contradiction. Thus, there exists some $P, Q \in \Syl_p(G)$ such that $[P : P \cap Q] \leq p^{e}$, as desired.

\[\tag*{$\blacksquare$}\]

Remark

I gave up on Part C during the exam – I think I wrote five different solutions, then a sad face. It was months later that I saw that it was an exercise in Dummit and Foote 😆 (Exercise 21, Section 6.2).

Problem 3

Suppose $A$ is a unital, commutative ring.

Part A

Suppose $I \trianglelefteq A$ and $a \in A$. Suppose $(I : \langle a \rangle) := \{r \in A \mid ra \in I\}$ and $I + \langle a \rangle$ are finitely generated ideals. Prove that $I$ is finitely generated.

Solution

Pick generators $x_1 + y_1 a, \ldots, x_n + y_n a$ for $I + \langle a \rangle$, where $x_i \in I, y_i \in A$, and generators $b_1, \ldots, b_k$ of $(I : \langle a \rangle)$. We claim that $I = \langle x_1, \ldots, x_n, b_1a, \ldots, b_ka \rangle$. It is clear that $\langle x_1, \ldots, x_n, b_1a, \ldots, b_ka \rangle \subseteq I$. On the other hand, let $z \in I$. Since $I \subseteq I + \langle a \rangle$, we obtain that there exist $c_1, \ldots, c_n \in A$ such that \begin{align*} z &= c_1(x_1 + y_1a) + \cdots + c_n (x_n + y_n a) \\ &= (c_1 x_1 + \cdots c_n x_n) + (c_1 y_1 + \cdots c_n y_n) a \end{align*}

So that

\begin{align*} \left(\sum_{i=1}^{n} c_iy_i\right)a = z - \sum_{i=1}^{n} c_ix_i \in I \end{align*}

Thus, $\sum_{i=1}^{n} c_iy_i \in (I : \langle a \rangle)$, so that $\left(\sum_{i=1}^{n} c_iy_i\right)a \in \langle b_1a, \ldots, b_ka\rangle$. Hence, $z \in \langle x_1, \ldots, x_n, b_1a, \ldots, b_ka \rangle$, thus $I = \langle x_1, \ldots, x_n, b_1a, \ldots, b_ka \rangle \subseteq I$, as desired.

Part B

Let $\Sigma := \{I \trianglelefteq A \mid I \text{ is not finitely generated}\}$. Prove that, if $\Sigma$ is not empty, then with respect to the inclusion ordering, $\Sigma$ has a maximal element.

Solution

We want to apply Zorn’s Lemma here, i.e. we want to show that any chain (totally ordered subset) in $\Sigma$ has an upper bound in $\Sigma$. Let $\{\aaa_i\}_{i \in I}$ be a chain in $\Sigma$. We claim that the union $\bigcup_{i \in I} \aaa_i$ is an upper bound in $\Sigma$. Via a routine argument, this is an ideal. To see that it is not finitely generated, suppose on the contrary, that $\bigcup_{i \in I} \aaa_i$ is finitely generated, say

\begin{equation*} \bigcup_{i \in I} \aaa_i = \langle a_1, \ldots, a_n \rangle \end{equation*}

Then, each $a_i$ is in some $\aaa_{\alpha_i}$, as it is in the union (and hence, $a_i \in \aaa_{\beta}$ for $\beta \geq \alpha_i$). Pick $\beta$ to be the greatest out of $\alpha_1, \ldots, \alpha_n$. Then,

\begin{equation*} \bigcup_{i \in I} \aaa_i = \langle a_{\alpha_1}, \ldots, a_{\alpha_n} \rangle \subseteq a_\beta \subseteq \bigcup_{i \in I} \aaa_i \end{equation*}

So, all the $\subseteq$ are $=$, and $\aaa_\beta$ is finitely generated, a contradiction to $\aaa_\beta \in \Sigma$. Thus, $\bigcup_{i \in I} \aaa_i$ is not finitely generated, so it is in $\Sigma$.

Part C

Suppose $I$ is a maximal element of $\Sigma$. Prove that $I$ is a prime ideal.

Solution

Suppose, on the contrary, that $I$ is not a prime ideal. Then, there exist $x, y \in A$ such that $xy \in I$, but $x,y \not\in I$. Since $I$ is maximal, $I + \langle x \rangle$ is finitely generated. Similarly, $I \subsetneq (I : \langle x \rangle)$ since $y \in (I : \langle x \rangle)$, so that $(I : \langle x \rangle)$ is finitely generated. Thus, by Part A, $I$ is finitely generated, a contradiction.

Remark

This problem essentially walks you through key steps in proving Cohen’s theorem on Noetherian rings (that a unital commutative ring $A$ is Noetherian if and only if all of its prime ideals are finitely generated). We showed it in homework for 200B, although I made a mistake when it was a homework exercise.

Problem 4

Suppose $A$ is a unital commutative ring and $I, J \trianglelefteq A$. Suppose $A/I$ is a flat $A$-module.

Part A

Prove that $I$ is a flat $A$-module.

Solution

Recall that if $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence of $A$-modules and $M_3$ is flat, then $M_2$ is flat $\iff$ $M_1$ is flat. And, recall that there exists the standard short exact sequence $0 \to I \embeds A \surjects A / I \to 0$. $A/I$ is flat by hypothesis, and $A$ is a flat $A$-module (as it is free). Thus, $I$ is flat.

Part B

Prove that there is an $A$-module isomorphism $\iota : I \tensor_A J → IJ$ such that $\iota(a \tensor b) = ab$ for every $a \in I$ and $b \in J$.

Solution

Consider the map $\hat{\iota}: I \times J \to IJ$ given by $(a, b) \mapsto ab$. It is easy to see that this is bilinear. Then, by the Universal Property of the Tensor Product, there is an $A$-module homomorphism $\iota: I \tensor_A J \to IJ$ given by $a \tensor b \mapsto ab$. To see that it’s surjective, notice that the image of $\iota$ contains a generating set of $IJ$. To see that it’s injective, note that tensoring the short exact sequence

\begin{equation*} 0 \to J \to A \to A / J \to 0 \end{equation*}

by $I$ gives that

\begin{equation*} 0 \to I \tensor J \to I \tensor A \to I \tensor A / J \to 0 \end{equation*}

is also exact. In particular, under the canonical identification of $I \tensor A \isom I$, we see that $I \tensor J \to I$ given by $a \tensor b \mapsto ab$ is injective, hence it restricts to an injective map $I \tensor J \to IJ$.

Part C

Prove that $IJ = I \cap J$.

Solution

In a general ring, we know that $IJ \subseteq I \cap J$ ($IJ$ is generated by elements of the form $ab$ with $a \in I, b \in J$, but such $ab \in I \cap J$). Let $x \in I \cap J$. Consider the exact sequence

\[\begin{CD} 0 @>>> I @>{id\mid_I}>> A @>{\pi}>> A/I @>>> 0\end{CD}\]

Tensoring this with $J$ and noting that $A/I$ is flat, we get the following exact sequence:

\[\begin{CD} 0 @>>> I \tensor J @>{id\mid_{I \tensor id}}>> A \tensor J @>{\pi \tensor id}>> A/I \tensor J @>>> 0\end{CD}\]

By Part B, $\iota: I \tensor J \to IJ$ via $a \tensor b \mapsto ab$ is an isomorphism. And, $\eta: A \tensor J \isom J$ via $a \tensor b \mapsto ab$ as well via the same argument as Part B. Finally, $A/I \tensor J \isom J/IJ$ via the map $(a + I) \tensor b \mapsto ab + IJ$. Indeed, the map $(A/I) \times J \to J/IJ$ given by $(a + I, b) \mapsto ab + IJ$ is well-defined (as the image of any two representatives $a + I$ differ by an element in $IJ$) and bilinear, hence it induces a homomorphism $\varphi: (A/I) \tensor J \to J/IJ$ given by $(a+I) \tensor b \mapsto ab + IJ$. On the other hand, $\psi: J \to (A/I) \tensor J$ given by $a \mapsto (1 + I) \tensor a$ is a well-defined homomorphism whose kernel contains $IJ$, so that $J/IJ \to (A/I) \tensor J$ is a well-defined homomorphism. One checks that $\varphi$ and $\psi$ are mutually inverse. We then get the diagram

\[\begin{CD} 0 @>>> I \tensor J @>{id\mid_{I \tensor id}}>> A \tensor J @>{\pi \tensor id}>> A/I \tensor J @>>> 0\\ @. @V{\iota}VV @V{\eta}VV @V{\psi}VV @. \\ 0 @>>> IJ @>{id\mid_{IJ}}>> J @>{\pi'}>> J/(IJ) @>>> 0\end{CD}\]

where $\pi': J \to J/(IJ)$ is the quotient map. One checks that this commutes. Since $x \in I \cap J$, $x \in J$, and $\eta^{-1}(x) = x \tensor 1$. But, $\psi((\pi \tensor id)(x \tensor 1)) = 0$, so by the commutativity of the diagram, $\pi'(x) = 0$, hence $x \in IJ$. So, $IJ = I \cap J$, as desired.

Remark

This was also on a previous qualifying exam, and I had actually seen it at 3 AM (but didn’t know how to solve it). I remember I thought I was really cooking on this problem, but after getting my qual back, I realized that the argument I wrote down was almost completely wrong for Parts B and C, and I only got 2/8 points for the problem. The solution presented here is from specializing the proof of a more general statement that we had done in 200B homework. Note that in the above, I used without proof the fact that if $N, M_1, M_2, M_3$ are $A$-modules, $M_3$ is flat, and $0 \to M_1 \to M_2 \to M_3 \to 0$ is a SES, then

\[\begin{CD} 0 @>>> M_1 \tensor N @>>> M_2 \tensor N @>>> M_3 \tensor N @>>> 0 \end{CD}\]

is also a SES, which we did for 200B homework.

Problem 5

Suppose $A$ is a Noetherian local ring and its only maximal ideal is $\mmm$.

Part A

Carefully state Nakayama’s lemma for $A$-modules.

Solution

If $M$ is a finitely generated $A$-module and $\mmm M = M$, then $M = 0$.

Part B

For every finitely generated $A$-module $M$, prove that \begin{equation*} d(M) = \dim_{A/\mmm}(M/\mmm M) \end{equation*} where $d(M)$ is the minimum number of elements needed to generate $M$.

Solution

It is clear that $d(M) \geq \dim_{A/\mmm}(M/\mmm M)$ since the cosets of any generating set of $M$ descend to a generating set of $M/\mmm M$. Let $n = \dim_{A / \mmm}(M / \mmm M)$. Then, there exist $m_1 + \mmm M, \ldots, m_n + \mmm M$ such that they generate $M/\mmm M$ as an $A/\mmm$ vector space. We claim that $m_1, \ldots, m_n$ generate $M$. Let $N$ be the submodule of $M$ generated by $m_1, \ldots, m_n$. By Part A, it is enough to show that $M/N = \mmm(M/N)$. It is clear that $\mmm(M/N) \subseteq M/N$. Let $x + N \in M/N$ (where $x \in M$). In $M/\mmm M$, we have $x + \mmm M = (a_1m_1 + \cdots + a_nm_n) + \mmm M$. Thus, \begin{equation*} x - (a_1m_1 + \cdots + a_nm_n) = y \in \mmm M \end{equation*} In particular, since $m_1, \ldots, m_n \in N$, we have that $x + N = y + N \in \mmm (M/N)$. Thus, $M / N \subseteq \mmm (M/N)$, so $M/N = \mmm (M/N)$. Thus, $M/N = 0$, hence $M = N$, so $d(M) \leq n = \dim_{A / \mmm M} (M / \mmm M)$, as desired.

Part C

Prove that every finitely generated projective $A$-module $P$ is free.

Solution

Since $P$ is finitely generated, we have the following short exact sequence

\[\begin{CD} 0 @>>> K @>>> A^{n} @>>> P @>>> 0 \end{CD}\]

where $n = d(P)$. By projectivity, $A^{n} \isom K \oplus P$. Then, we have that \begin{equation*}
(A/\mmm)^n \isom (K/\mmm K) \oplus (P / \mmm P) \end{equation*} as $A/\mmm$ vectorspaces. By Part B, $\dim_{A / \mmm} (P/\mmm P) = n$, so that $\dim_{A / \mmm} \frac{K}{\mmm K} = 0$. By Part B, $d(K) = 0$, so $K = 0$. Thus, $P \isom A^{n}$ is free, as desired.

Remark

This was a problem from 200B homework.

Problem 6

Suppose $F$ is a field of characteristic zero and $f$ is an irreducible polynomial of degree $n$ in $F[x]$. Let $E$ be a splitting field of $f$ over $F$. Let $\{\alpha_1, \ldots, \alpha_n\}$ be the set of zeros in $f$ in $E$.

Part A

Prove that if $\Gal(E/F)$ is abelian, then $E = F[\alpha_i]$ for every $i$.

Solution

Since $\Gal(E / F)$ is abelian, we see that $F[\alpha_i] / F$ is Galois, as any subfield of an extension with abelian Galois group is Galois by the main theorem of Galois theory: $\Gal(E/F[\alpha_i])$ is a normal subgroup of $\Gal(E / F)$ (as any subgroup of an abelian group is normal), so $F[\alpha_i]/F$ is normal and separable (since $F[\alpha_i]$ is characteristic zero). Hence, $F[\alpha_i]/F$ is Galois.

But, as $F[\alpha_i]$ is normal and $f$ has a root in $F[\alpha_i]$, $f$ splits in $F[\alpha_i]$, so $F[\alpha_i]$ is the splitting field of $f$, thus $F[\alpha_i] = E$.

Part B

Prove that if $n = p$ is prime, then $F[\alpha_i]$ implies that $\Gal(E/F) \isom \Z/p\Z$.

Solution

\begin{equation*} \left|\Gal(E/F)\right| = \left|\Gal(F[\alpha_i]/F)\right| = [F[\alpha_i] : F] = p \end{equation*}

Thus, $\Gal(E / F) \isom \Z/p\Z$.

Remark

This problem was on the extremely easy side for a qualifying exam, not sure what happened here.

Problem 7

Let $K$ be a field with 729 elements. Let $\F_3$ be the prime field of $K$ and set $G = \Gal(K / \F_3)$. Consider the action of $G$ on $K$ given by $\sigma \cdot u = \sigma(u)$. Describe the orbits of this action, calculate their sizes, and calculate how many orbits are there of each size.

Solution

Note that $K = \F_{3^6}$. By the Galois theory of finite fields, we know $G \isom C_6 = \langle \sigma \rangle$, with $\sigma: \F_{3^6} \to \F_{3^6}$ via $x \mapsto x^3$ being the Frobenius. The subfields of $K$ are $\F_{3^{3}}$, $\F_{3^{2}}$, and $\F_{3}$, which are Galois over $\F_{3}$ (and $\F_{3^{3}} \cap \F_{3^{2}} = \F_{3}$). Let $u \in K$. If $u \in \F_{3^3}$, then $\sigma \cdot u \in \F_{3^3}$ and $u \in \F_{3^2}$, then $\sigma \cdot u \in \F_{3^2}$. This is since $\sigma$ must send $u \in \F_{3^3}$ to another root of $x^{3^{3}}-x$ and $u \in \F_{3^2}$ to another root of $x^{3^2}-x$ (and $\F_{3^3}$ and $\F_{3^2}$ are the splitting fields of these polynomials). Also, $\sigma$ must fix $\F_{3}$. Putting this all together, we see that:

If $u \in \F_{3}$, then $\sigma \cdot u = u$, the orbit of $u \in \F_{3}$ is $\{u\}$.

If $u \in \F_{3^2} \setminus \F_3$, note that $\sigma$ is injective and $\sigma \mid_{\F_{3^2}}$ has order $2$ in $\Gal(\F_{3^2}/\F_3)$, so that the orbit of $u$ is $\{u, \sigma(u)\}$.

If $u \in \F_{3^3} \setminus \F_3$, note that $\sigma$ is injective and $\sigma \mid_{\F_{3^2}}$ has order $3$ in $\Gal(\F_{3^3}/\F_3)$, so that the orbit of $u$ is $\{u, \sigma(u), \sigma^2(u)\}$.

If $u \in \F_{3^6} \setminus (\F_3 \cup \F_{3^2} \cup \F_{3^3})$, note that $\sigma$ is injective and $\sigma \mid_{\F_{3^6}}$ has order $3$ in $\Gal(\F_{3^3}/\F_3)$, so that the orbit of $u$ is $\{u, \sigma(u), \sigma^2(u), \ldots, \sigma^{5}(u)\}$.

Thus, there are:

$3$ orbits of size $1$
$\frac{3^2-3}{2} = 3$ orbits of size $2$
$\frac{3^3-3}{3} = 8$ orbits of size $3$
$\frac{3^6-(3^3-3)-(3^2-3)-3}{6} = 116$ orbits of size $6$

Conclusion

I will definitely say that I was quite lucky in the whole ordeal, especially in actually being able to do well. I did not end up doing much preparation for the exam, as the midterm date for my undergrad Real Analysis got rescheduled to the day before the qual at the last minute! So, the only studying I ended up doing was from 2-4 AM on my bed the night before (morning of?). I ended up just reading over all my homework solutions for the past 3 quarters, and took very cursory glances at all the qualifying exams over the past 5 or so years, hoping that the ones I couldn’t solve wouldn’t show up. All in all, I am fairly happy with my result ($45/56 \approx 80.3\%$), and I think it makes a pretty nice cherry on top of my math journey at UCSD.