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Consider two matrices $A$ and $B$,

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & -1 \end{bmatrix},\ B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 &0 & -1 \end{bmatrix}.$$

Let $M,N$ be the $\mathbb{R}[x]$-modules defined by $A$ and $B$, respectively. Then is $M$ isomorphic to $N$? What are submodules of $M$?

In $M$, for any $v\in \mathbb{R}^3$, $xv \mapsto Av$. Hence $x^2v\mapsto v$, $x^3v \mapsto Av$ ... Then elements in $M$ take the form $a_1Av+a_2v$, where $a_1,a_2\in\mathbb{R}$. Similarly elements in $N$ take the form $b_1Av+b_2v$.

Then I construct a homomorphism $\phi:M\to N$, $a_1Av+a_2v\mapsto a_1Bv+b_2v$. I have verified that $\phi(m_1+m_2)=\phi(m_1)+\phi(m_2)$ and $\phi(p(x)m_1)=p(x)\phi(x_1)$. $\phi$ is surjective clearly. However, I cannot prove $\phi$ is injective.

Also, I have no clue finding the submodule of $M$.

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  • $\begingroup$ What do you mean by the submodule of $M$? $\endgroup$ Feb 19, 2017 at 16:34
  • $\begingroup$ @AndreasCaranti by submodule I mean "subset of $M$ which is closed under addition and scalar multiplication". Sorry I use "the". Actually I think there are many submodules. $\endgroup$
    – Kenneth.K
    Feb 19, 2017 at 16:39
  • $\begingroup$ My problem was indeed with the definite article, as it denotes uniqueness. $\endgroup$ Feb 19, 2017 at 16:40

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$\renewcommand{\phi}{\varphi}$$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Suppose $\phi : M \to N$ is an isomorphism. If $(x - 1) v = 0$, for $v \in M$, then, we will have $$0 = \phi((x - 1) v) = (x - 1) \phi(v).$$ The converse also holds, as $\phi$ is an isomorphism.

But note that $$ \Set { v \in M : (x - 1) v = 0} $$ has dimension $2$ over $\R$, whereas $$ \Set { w \in N : (x - 1) w = 0} $$ has dimension $1$.

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  • $\begingroup$ "If $(x-1)v=0$, for $x\in M$", do you mean $x\in \mathbb{R}[x]$? $\endgroup$
    – Kenneth.K
    Feb 19, 2017 at 16:41
  • $\begingroup$ Sorry, it was a typo. Just fixed, I meant $v \in M$. $\endgroup$ Feb 19, 2017 at 16:43
  • $\begingroup$ Why $\{v\in M:(x-1)v=0\}$ has dimension 2 over $\mathbb{R}$? $(x-1)v=0$ is equivalent to $Av=v$ right? $\endgroup$
    – Kenneth.K
    Feb 19, 2017 at 16:45
  • $\begingroup$ Yes, just compute $A - I$ and $B - I$ and you'll see what I mean. $\endgroup$ Feb 19, 2017 at 16:46
  • $\begingroup$ For submodules of $M$, I guess they are the subgroup of $M$ generated by $(a,b,c)$ and $(a,b,-c)$ for any $a,b,c \in \mathbb{R}$? $\endgroup$
    – Kenneth.K
    Feb 19, 2017 at 16:54

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