# Abstract Nonsense

## Halmos Section 21 and Section 22:Quotient Spaces and Dimension of a Quotient Space

Point of post: In this post I will do the problems from sections 21 and 22 in Halmos. There are some quite interesting ones, with some definite algebraic slants.

1.

Problem: Consider the quotient spaces obtained by reducing the space $\mathbb{C}[x]$ modulo various subspaces. If $\mathscr{M}=\mathscr{P}_n$, the set of all $n$th degree polynomials, is $\mathbb{C}[x]/\mathscr{P}_n$ finite-dimensional? What if $\mathscr{M}$ is the set of all even polynomials. What if $\mathscr{M}$ consists of all polynomials divisible by $x^n$?

Proof: When $\mathscr{M}=\mathscr{P}_n$ then $\mathbb{C}[x]/\mathscr{M}$ is not finite dimensional. To see this fix $m_0$ and consider $x^{n+1}+\mathscr{M},\cdots,x^{n+m_0+1}\mathscr{M}$ suppose then that

$\displaystyle \sum_{j=1}^{m_0+1}\alpha_j\left(x^{n+j}+\mathscr{M}\right)=\mathscr{M}$

but, by definition this implies that

$\displaystyle \sum_{j=1}^{m_0+1}\alpha_jx^{n+j}+\mathscr{M}=\mathscr{M}$

But, this is the same as saying that

$\displaystyle \sum_{j=1}^{m_0+1}\alpha_jx^{n+j}\in\mathscr{M}$

which is clearly impossible unless $\alpha_1=\cdots=\alpha_{m_0+1}=0$. Thus, $\left\{x^{n+1}+\mathscr{M},\cdots,x^{n+m_0+1}+\mathscr{M}\right\}$ are $m_0+1$ linearly independent vectors in $\mathbb{C}/\mathscr{M}$ and so it can’t be $m_0$-dimensional. Since $m_0$ was arbitrary it follows that $\dim_\mathbb{C}\mathbb{C}[x]/\mathscr{M}\ne n$ for any $n\in\mathbb{N}$ and is thus not finite dimensional.

Now, if $\mathscr{M}$ is the set of all even polynomials, then since $\mathbb{C}[x]=\mathscr{M}\oplus\mathscr{N}$ where $\mathscr{N}$ is the set of all odd polynomials we know that $\mathbb{C}[x]/\mathscr{M}\cong\mathscr{N}$ and since $\mathscr{N}$ isn’t finite dimensional (consider $x,x^3,x^5,\cdots$) we may conclude that $\mathbb{C}[x]/\mathscr{M}$ is not finite.

Lastly, we claim that if $\mathscr{M}$ is the set of all polynomials divisible by $x^n$ that $\left\{1+\mathscr{M},\cdots,x^{n-1}+\mathscr{M}\right\}$ is a basis for $\mathbb{C}[x]/\mathscr{M}$. To see this first suppose that

$\displaystyle \sum_{j=0}^{n-1}\alpha_j \left(x^j+\mathscr{M}\right)=\sum_{j=0}^{n-1}\alpha_j x^j+\mathscr{M}=\mathscr{M}$

then, by definition

$\displaystyle \sum_{j=0}^{n-1}\alpha_j x^j\in\mathscr{M}$

which is evidently impossible unless $\alpha_0=\cdots=\alpha_{n-1}=0$. Next, let $p(x)+\mathscr{M}\in\mathbb{C}[x]/\mathscr{M}$ be arbitrary. We know that $\displaystyle p(x)=\sum_{j=0}^{m}\alpha_j x^j$ for some $m\in\mathbb{N}$. If $m\leqslant n-1$ the result’s clear, and if not we see that

\displaystyle \begin{aligned}p(x)+\mathscr{M} &= \sum_{j=0}^{m}\alpha_j x^j+\mathscr{M}\\ &= \left(\sum_{j=0}^{n-1}\alpha_j x^j+\sum_{j=n}^{m}\alpha_j x^j\right)+\mathscr{M}\\ &= \bigg(\sum_{j=0}^{n-1}\alpha_j x^j+\mathscr{M}\bigg)+\bigg(\overbrace{\sum_{j=n}^{m}\alpha_j {x^j}}^{\text{in }\mathscr{M}}+\mathscr{M}\bigg)\\ &=\sum_{j=0}^{n-1}\alpha_j x^j+\mathscr{M}\end{aligned}

from where it follows that $\text{span }\left\{1+\mathscr{M},\cdots,x^{n-1}+\mathscr{M}\right\}=\mathbb{C}[x]/\mathscr{M}$. Thus, $\dim_{\mathbb{C}}\mathbb{C}[x]/\mathscr{M}=n$

2.

Problem: IF $\mathscr{S}$ and $\mathscr{J}$ are arbitrary subsets of a vector space (not necessarily cosets of a subspace), there is nothing to stop us from defining $\mathscr{S}+\mathscr{J}$ just as addition was defined for cosets, and similarly, we may define $\alpha\mathscr{S}$. IF the class of all subsets of a vector space is endowed with these “linear operations” which of the axioms of a vector space are satisfies?

Proof: The following uses the definition of a field as is given on page three of Halmos’s book:

A)

1. This axiom (commutativity of the addition) is definitely true since

$\mathscr{S}+\mathscr{J}=\left\{s+j:s\in\mathscr{S}\text{ and }j\in\mathscr{J}\right\}=\left\{j+s:j\in\mathscr{J}\text{ and }s\in\mathscr{S}\right\}=\mathscr{J}+\mathscr{S}$

2. This is also true since

\begin{aligned}\mathscr{S}+\left(\mathscr{J}+\mathscr{Q}\right) &=\left\{s+k:s\in\mathscr{S}\text{ and }k\in\mathscr{J}+\mathscr{Q}\right\}\\ &= \left\{s+(j+q):s\in\mathscr{S}\text{ and }j\in\mathscr{J}\text{ and }q\in\mathscr{Q}\right\}\\ &= \left\{(s+j)+q:s\in\mathscr{S}\text{ and }j\in\mathscr{J}\text{ and }q\in\mathscr{Q}\right\}\\ &=\left\{k+q:k\in\mathscr{S}+\mathscr{J}\text{ and }q\in\mathscr{Q}\right\}\\ &= \left(\mathscr{S}+\mathscr{J}\right)+\mathscr{Q}\end{aligned}

3. This is true since $\{e\}+\mathscr{S}=\mathscr{S}+\{e\}=\left\{e+s:s\in\mathscr{S}\right\}=\left\{s:s\in\mathscr{S}\right\}=\mathscr{S}$

4. This is not true. Let $\mathbb{R}$ be a vector space over itself and consider the subset $\mathbb{Z}$. Note then that since $0\in\mathbb{Z}$ we must have that $s\in\mathbb{Z}+\mathscr{S}$ for all $s\in\mathscr{S}$. So, in particular if $\mathbb{Z}+\mathscr{J}=\{e\}$ then it must be true that $\mathscr{J}=\{e\}$, but this contradicts that $\mathbb{Z}+\mathscr{J}=\{e\}$.

B)

1. This is true since $\alpha\left(\beta\mathscr{J}\right)=\alpha\left\{\beta j:j\in\mathscr{J}\right\}=\left\{\alpha(\beta j):j\in\mathscr{J}\right\}=\left\{(\alpha\beta)j:j\in\mathscr{J}\right\}=(\alpha\beta)\mathscr{J}$

2. This is also true since $1\mathscr{J}=\left\{1j:j\in\mathscr{J}\right\}=\left\{j:j\in\mathscr{J}\right\}=\mathscr{J}$

3. This is true since

\begin{aligned}\alpha\left(\mathscr{S}+\mathscr{J}\right) &=\alpha\left\{s+j:s\in\mathscr{S}\text{ and }j\in\mathscr{J}\right\}\\ &=\left\{\alpha(s+j):s\in\mathscr{S}\text{ and }j\in\mathscr{J}\right\}\\ &=\left\{\alpha s+\alpha j:s\in\mathscr{S}\text{ and }j\in\mathscr{J}\right\}\\ &=\left\{k+\ell:k\in\alpha\mathscr{S}\text{ and }\ell\in\alpha\mathscr{J}\right\}\\ &=\alpha\mathscr{S}+\alpha\mathscr{J}\end{aligned}

4.This is true since

\begin{aligned}(\alpha+\beta)\mathscr{S} &=\left\{(\alpha+\beta)s:s\in\mathscr{S}\right\}\\ &=\left\{\alpha s+\beta s:s\in\mathscr{S}\right\}\\ &=\left\{k+\ell:k\in\alpha\mathscr{S}\text{ and }\ell\in\beta\mathscr{S}\right\}\\ &=\alpha\mathscr{S}+\beta\mathscr{S}\end{aligned}

Problem:

a) Suppose that $\mathscr{M}$ is a subspace of a vector space $\mathscr{V}$. Two vectors $x$ and $y$ of $\mathscr{V}$ are congruent modulo $\mathscr{M}$, denoted $x\equiv y\text{ mod }\mathscr{M}$, if $x-y\in\mathscr{M}$. Prove that $\equiv\text{ mod }\mathscr{M}$ is an equivalence relation on $\mathscr{V}$

b) If $\alpha_1,\alpha_2\in F$ and if $x_1,y_1,x_2,y_2\in\mathscr{V}$ are such that $x_1\equiv y_1\text{ mod }\mathscr{M}$ and $x_2\equiv y_2\text{ mod }\mathscr{M}$ prove that $\alpha_1x_1+\alpha_2x_2\equiv \alpha_1y_1+\alpha_2y_2\text{ mod }\mathscr{M}$.

c) Prove that the equivalence classes of $\equiv\text{ mod }\mathscr{M}$ are precisely the cosets of $\mathscr{M}$.

Proof:

a) Clearly $x\equiv x\text{ mod }\mathscr{M}$ since $x-x=\bold{0}\in\mathscr{M}$. Also, if $x\equiv y\text{ mod }\mathscr{M}$ then $x-y\in\mathscr{M}$ and thus $-1(x-y)=y-x\in\mathscr{M}$ and so $y\equiv x\text{ mod }\mathscr{M}$. Lastly, if $x\equiv y\text{ mod }\mathscr{M}$ and $y\equiv z\text{ mod }\mathscr{M}$ then $x-y\in\mathscr{M}$ and $y-z\in\mathscr{M}$ and thus $(x-y)+(y-x)=x-z\in\mathscr{M}$ so that $x\equiv z\text{ mod }\mathscr{M}$

b) Clearly this is true since $x_1-y_1\in\mathscr{M}$ and $x_2-y_2\in\mathscr{M}$ implies that $\alpha_1(x_1-y_1)-\alpha_2(x_2-y_2)=(\alpha_1x_1-\alpha_2x_2)-(\alpha_1y_1-\alpha_2y_2)\in\mathscr{M}$ so that $\alpha_1x_1-\alpha_2x_2\equiv\alpha_1y_1-\alpha_2y_2\text{ mod }\mathscr{M}$.

c) Let $[x]=\left\{y\in\mathscr{V}:x\equiv y\text{ mod }\mathscr{M}\right\}$, we claim that $[x]=x+\mathscr{M}$. To see this let $y\in[x]$ then $x-y\in\mathscr{M}$ so that $x-y=m$ for some $m\in\mathscr{M}$. Thus, $y=x-m$ and thus $y\in x+\mathscr{M}$. Conversely, if $y\in x+\mathscr{M}$ then $y=x+m$ for some $m\in\mathscr{M}$ and so $x-y=-m\in\mathscr{M}$ so that $x\equiv y\text{ mod }\mathscr{M}$ and thus $y\in[x]$.

3.

Problem:

a) Suppose that $\mathscr{M}$ is a subspace of a vector space $\mathscr{V}$. Prove that $\text{Hom}\left(\mathscr{V}/\mathscr{M},F\right)\cong\text{Ann }\mathscr{M}$ without appealing to dimensionality.

b) Show that $\text{Hom}\left(\mathscr{V},F\right)/\left(\text{Ann }\mathscr{M}\right)\cong\text{Hom}\left(\mathscr{M},F\right)$ without appealing to dimensionality.

Proof:

a) Let $\{x_1,\cdots,x_m\}$ be a basis for $\mathscr{M}$ and extend it to a basis $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$. Then, we know that $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ lifts to the dual basis $\{\varphi_{x_1},\cdots,\varphi_{x_m},\varphi_{y_1},\cdots,\varphi_{y_k}\}$ for $\text{Hom}\left(\mathscr{V},F\right)$, with, in particular, $\{\varphi_{y_1},\cdots,\varphi_{y_k}\}$ be a basis for $\text{Ann }\mathscr{M}$.

But, with equal verity we may start with $\{x_1,\cdots,x_m,y_1,\cdots,y_k\}$ and produce the basis $\left\{y_1+\mathscr{M},\cdots,y_k+\mathscr{M}\right\}$ for $\mathscr{V}/\mathscr{M}$. And from there we may lift to the basis $\left\{\varphi_{y_1+\mathscr{M}},\cdots,\varphi_{y_k+\mathscr{M}}\right\}$ for $\text{Hom}\left(\mathscr{V}/\mathscr{M},F\right)$.

From here the course of action is clear, namely let

$f:\{\varphi_{y_1},\cdots,\varphi_{y_k}\}\to\left\{\varphi_{y_1+\mathscr{M}},\cdots,\varphi_{y_k+\mathscr{M}}\right\}:\varphi_{y_\ell}\mapsto\varphi_{y_\ell+\mathscr{M}}$

and extend it to the map

$f^{*}:\text{Ann }\mathscr{M}\to\text{Hom}\left(\mathscr{V}/\mathscr{M},F\right)$

by linearity. Clearly, by construction, we see that $f^{*}$ is linear. To see it’s injective suppose that

$\displaystyle f\left(\sum_{j=1}^{k}\alpha_j \varphi_{y_j}\right)=\sum_{j=1}^{k}\alpha_j \varphi_{y_j+\mathscr{M}}=\bold{0}$

then, since $\left\{\varphi_{y_1+\mathscr{M}},\cdots,\varphi_{y_k+\mathscr{M}}\right\}$ is linearly independent we can see that $\alpha_1=\cdots=\alpha_k=0$. Thus, $\sum_{j=1}^{k}\alpha_j\varphi_{y_j}=\bold{0}$. Also, it’s clear that $f$ is surjective since if $\displaystyle \sum_{j=1}^{k}\alpha_j\varphi_{y_j+\mathscr{M}}$ is in $\text{Hom}\left(\mathscr{V}/\mathscr{M},F\right)$ we have that $\displaystyle \sum_{j=1}^{k}\alpha_j \varphi_{y_j}\in\text{Ann }\mathscr{M}$ and $\displaystyle f\left(\sum_{j=1}^{k}\alpha_j \varphi_{y_j}\right)=\sum_{j=1}^{k}\alpha_j\varphi_{y_j+\mathscr{M}}$

b) Consider the mapping

$f:\text{Hom}\left(\mathscr{V},F\right)/\left(\text{Ann }\mathscr{M}\right)\to\text{Hom}\left(\mathscr{M},F\right):\varphi+\text{Ann }\mathscr{M}\to \varphi_{\mid\mathscr{M}}$

Now, linearity is clear since

\begin{aligned} f\left(\left(\varphi_1+\text{Ann }\mathscr{M}\right)+\left(\varphi_2+\text{Ann }\mathscr{M}\right)\right) &= f\left((\varphi_1+\varphi_2)+\text{Ann }\mathscr{M}\right)\\ &=\left(\varphi_1+\varphi_2\right)_{\mid\mathscr{M}}\\ &= \left(\varphi_1\right)_{\mid\mathscr{M}}+\left(\varphi_2\right)_{\mid\mathscr{M}}\\ &= f\left(\varphi_1+\text{Ann }\mathscr{M}\right)+f\left(\varphi_2+\text{Ann }\mathscr{M}\right)\end{aligned}

and

$f\left(\alpha\left(\varphi+\text{Ann }\mathscr{M}\right)\right)=f\left((\alpha\varphi)+\text{Ann }\mathscr{M}\right)=\left(\alpha\varphi\right)_{\mid\mathscr{M}}=\alpha\left(\varphi\right)_{\mid\mathscr{M}}=\alpha f\left(\varphi+\text{Ann }\mathscr{M}\right)$

Now, to prove injectivity suppose that $\varphi_1+\text{Ann }\mathscr{M}\ne\varphi_2+\text{Ann }\mathscr{M}$. Then, there exists some $m_0\in\mathscr{M}$ such that $\varphi_1(m_0)\ne\varphi_2(m_0)$. Otherwise, for all $m\in\mathscr{M}$ we’d have that $\varphi_1(m)=\varphi_2(m)\implies \varphi_1(m)-\varphi_2(m)=\bold{0}$ and so $\varphi_1-\varphi_2\in\text{Ann }\mathscr{M}$, but this (by looking at problem 2.) says that $\varphi_1+\text{Ann }\mathscr{M}=\varphi_2+\text{Ann }\mathscr{M}$ which we assumed otherwise. Thus, there exists the aforementioned $m_0$ and so $(\varphi_1)_{\mid\mathscr{M}}\ne(\varphi_2)_{\mid\mathscr{M}}$

Lastly, to prove surjectivity let $\mathscr{N}\subseteq\mathscr{V}$ be such that $\mathscr{V}=\mathscr{M}\oplus\mathscr{N}$. Then, we may extend any $\varphi\in\text{Hom}\left(\mathscr{M},F\right)$ to a $\varphi^*:\mathscr{V}\to F$ by defining $\varphi^*(x)=\varphi(m)$ where $m+n$ is the unique representation of $x$ as the sum of an element of $\mathscr{M}$ and $\mathscr{N}$. Clearly then $\varphi^*+\text{Ann }\mathscr{M}\in\text{Hom}\left(\mathscr{V},F\right)/\left(\text{Ann }\mathscr{M}\right)$ and $\varphi^*+\text{Ann }\mathscr{M}\overset{f}{\longmapsto}\varphi$. Thus, $f$ is an isomorphism.

October 18, 2010 -

1. I would like to thank you for posting these nice proofs.
Do you have also a proof for problem 2 section 22? The number 2 from this post is actually problem 3 from the same section 22.
Thank you in advance! Any help will be appreciated.

Comment by Kira | November 19, 2010 | Reply

• There you go. This is the second time I’ve accidentally skipped writing up a solution. If you find any more let me know!

Comment by drexel28 | November 19, 2010 | Reply

Comment by Kira | November 19, 2010 | Reply

3. 1.I would like to thank you for posting these nice proofs.
Do you have also a proof for problem 2 section 22? The number 2 from this post is actually problem 3 from the same section 22.
Thank you in advance! Any help will be appreciated.

Comment by Kira | November 19, 2010 | Reply

• I’m not sure what you mean. I just did the problem I missed, above.

Comment by drexel28 | November 21, 2010 | Reply

• Thank you! I really appreciate your great work!

Comment by Kira | November 21, 2010

• You’re quite welcome!

Comment by drexel28 | November 22, 2010