Solutions to the Final Exam for MATH2510: Algebra, Fall 2024, Brown University
Problem 1. Let $D_{2n}$ be the symmetry group of an $n-$gon and $S_n$ be the permutation group of $n$ elements. Determine whether the following two statements are true:
- $D_{12}\cong D_6\times S_2$
- $D_{16}\cong D_8\times S_2$
Proof: We present here only a geometric argument for the first statement. $D_{12}$ is not only the symmetry group of a hexagon but also the one of a hexagram, which can be decomposed into two triangles. The figure below shows two elements of $D_{12}$. The first row shows how you can identify the 60 degree rotation of a triangle with the 120 degree rotation of the hexagram. The second row shows how you can identify the transposition in $S_2$ with a reflection of the hexagram.
$D_{16}$ is not isomorphic to $D_8\times S_2$ because the first group contains an element of order 8 and the degree of every element in the second group is less than or equal to 4. $\square$
Problem 2. Let $G=\text{SL}_2(\mathbb{F}_3)$ and $H\subset G$ be the group of upper triangular matrices with all ones on the diagonal. Show that $H$ is a Sylow 3-subgroup of $G$. Compute the number of Sylow 3-subgroups of $G$.
Proof: Clearly we have
\[\begin{align*} H&=\{\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix}\}\\ |H|&=3\\ H&\subset G\subset \text{GL}_2(\mathbb{F}_3) \end{align*}\]Hence, $3\lvert \lvert G\lvert$ and $\lvert G\lvert\lvert \lvert\text{GL}_2(\mathbb{F}_3)\lvert$. Using the formula mentioned in the practice final, we have
\[|\text{GL}_2(\mathbb{F}_3)|=(3^2-1)(3^2-3)=48=3\times 2^4\]Therefore, the largest integer $m$ such that $3^m$ divides $\lvert G\rvert$ is $1$, which implies that $H$ is a Sylow 3-subgroup of $G$.
Now we compute the number $n$ of Sylow 3-subgroups. We first notice that the group of lower triangular matrices with all ones on the diagonal, denoted by $L$, is a subgroup of $G$. $L$ also contains 3 elements and is disjoint from $H$. Moreover,
\[\begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}\notin G\]so we have the following estimates:
\[6\leq |G|<48\]The only possible choices for $|G|$ are $3\times 2, 3\times 2^2, 3\times 2^3$. By the third Sylow theorem, $n\mathrel{|}\cfrac{|G|}{3}$. $n$ thus divides $8$, so the only possible choices for $|G|$ are $1,2,4,8$. The third Sylow theorem also says that $n\equiv 1\text{ (mod }3)$, which implies that $n$ is either equal to $1$ or $4$. Since $L$ is also a Sylow 3-subgroup, $n\geq 2$. We conclude that $n=4$. $\square$
Problem 3. VII.6.7. (Algebra: Chapter 0) Let $R$ be an integral domain and a nonzero element $a$. Give an example such that $a$ is irreducible but not prime.
Proof: Let $R=\cfrac{\mathbb{C}[x,y,z,w]}{(xy-zw)}$. $x+(xy-zw)$ is irreducible. The ideal generated by $x+(xy-zw)$ in $R$ contains $zw+(xy-zw)=xy+(xy-zw)$, but it does not contain $z+(xy-zw)$ and $w+(xy-zw)$, which means that $x+(xy-zw)$ is not prime. $\square$
Problem 4. Prove that every nonzero commutative ring contains at least one maximal ideal.
Proof: This is just a classical application of Zorn’s lemma. $\square$
Problem 5. Prove that a right adjoint functor preserves monomorphisms.
Proof: Given two functors $F:\mathcal{C}\rightarrow \mathcal{D}$ and $G:\mathcal{D}\rightarrow \mathcal{C}$. If $G$ is right adjoint to $F$, then we have the following bijection:
\[\hom_{\mathcal{D}}(FX,Y)\overset{I_{X,Y}}{\cong} \hom_{\mathcal{C}}(X,GY)\]for all objects $X$ in $\mathcal{C}$ and $Y$ in $\mathcal{D}$. Moreover, the following diagram commutes:
where $f:Y\rightarrow Y’$ is a morphism in $\mathcal{D}$. Now we assume that $f$ is a monomorphism. We want to prove that $Gf$ is a monomorphism. Suppose $g,h: X\rightarrow GY$ are two morphisms such that
\[Gf\circ g=Gf\circ h\]Our goal is to show that $g=h$. Since $I_{X,Y}$ is a bijection, there exists morphisms $a,b:FX\rightarrow Y$ such that
\[\begin{align*} g&=I_{X,Y}(a)\\ h&=I_{X,Y}(b) \end{align*}\]We then deduce that
\[\begin{align*} Gf\circ g&=Gf\circ h\\ Gf\circ I_{X,Y}(a)&=Gf\circ I_{X,Y}(b)\\ I_{X,Y'}(f\circ a)&=I_{X,Y'}(f\circ b)\text{ (because the diagram commutes)}\\ f\circ a&=f\circ b\text{ (because $I_{X,Y'}$ is injective)}\\ a&=b\text{ (because $f$ is a monomorphism)}\\ I_{X,Y}(a)&=I_{X,Y}(b)\\ g&=h \end{align*}\]$\square$
Problem 6. Let $A:\mathbb{C}^n\rightarrow \mathbb{C}^n$ be an invertible map of finite order ($A^k=I$ for some $k\geq 0$). Show that $A$ is diagonalizable.
Proof: Over $\mathbb{C}$, a matrix is diagonalizable iff its minimal polynomial is separable. In our case, the minimal polynomial $m_A(x)$ of $A$ divides $x^k-1$. Since $x^k-1$ has no repeated roots, $m_A(x)$ also has no repeated roots, which implies that $A$ is diagonalizable. $\square$
Problem 7. Is there an epimorphism in the category of fields that is not an isomorphism?
Proof: There indeed exists an epimorphism that is not an isomorphism in the category of fields. Let $p$ be a prime. Consider the field of rational functions $\mathbb{F}_p(t)$ and a simple extension $\mathbb{F}_p(t)(t^{\frac{1}{p}})$. Then the inclusion map $i:\mathbb{F}_p(t)\rightarrow \mathbb{F}_p(t)(t^{\frac{1}{p}})$ is clearly not surjective and thus not an isomorphism. We claim that this is an epimorphism. We need to show that if $g,h$ are two morphisms from $\mathbb{F}_p(t)(t^{\frac{1}{p}})$ to a field $k$ such that
\[g\circ i=h\circ i\]then $g=h$. It suffices to prove that $g(t^{\frac{1}{p}})=h(t^{\frac{1}{p}})$. Let $m(x)=x^p-g(t)\in k[x]$. We have
\[\begin{align*} (x-g(t^{\frac{1}{p}}))^p&=x^p-(g(t^{\frac{1}{p}}))^p\\ &=x^p-g(t)\\ &=x^p-g\circ i(t)\\ &=x^p-h\circ i(t)\\ &=x^p-h(t)\\ &=x^p-(h(t^{\frac{1}{p}}))^p\\ &=(x-h(t^{\frac{1}{p}}))^p \end{align*}\]If $g(t^{\frac{1}{p}})$ is not equal to $h(t^{\frac{1}{p}})$, then we have two different factorizations into irreducibles of the polynomial $m(x)$, which is a contradiction to the fact that $k[x]$ is a UFD. $\square$
Problem 8. Let $k$ be a field such that its characteristic is not equal to $2$. Consider an extension $k\subset L$ of degree 2. Show that $L$ is a Galois extension.
Proof: Take an element $\alpha\in L-k$. Clearly $1<[k(\alpha):k]\leq 2$, which implies that $[k(\alpha):k]=2$ and $L=k(\alpha)$. $\alpha$ is then algebraic, and its minimal polynomial $m(x)$ is quadratic. $m(x)$ then splits in $L$ and has the following form for some $\beta\in L$:
\[m(x)=(x-\alpha)(x-\beta)\]$L=k(\alpha)\subset k(\alpha,\beta)\subset L$, so $L$ is the splitting field of $m$. Suppose $\beta=\alpha$, then
\[m(x)=x^2-2\alpha x+\alpha^2\]Since the characteristic of $k$ is not equal to $2$ and $2\alpha\in k$, $\alpha=\frac{1}{2}(2\alpha)\in k$. This contradiction implies that $\beta\ne \alpha$ and thus $m$ is a separable polynomial. $L$ is then Galois because it is the splitting field of a separable polynomial. $\square$