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Abstract Algebra - Part II

From DF, Section 10.3, p. 356ff. Note that $R$ is assumed to have a unit element.

Problem 1

Prove that if $A$ and $B$ are sets of the same cardinality then $F(A)$ and $F(B)$ are isomorphic.

Problem 2

Prove that, if $R$ is commutative, then $R^{n}$ and $R^{k}$ are isomorphic if and only if $n=k$.

Strategy:

Assume $f:R^{n}\to R^{k}$ is an isomorphism.

  1. If $R$ is a field, this follows from the fact that the dimension of a vector space is well-defined.
  2. If $R$ is not a field, it has a proper maximal ideal $m$ so that $R/m$ is a field.
  3. $R^{n}/mR^{n}=(R/m)^{n}$. Any elememt $(mr_1, mr_2,\ldots, mr_n)$ belongs to $m^{n}\subset R^{n}$. Conversely, if $x=(m_1,\ldots, m_n)\in m^{n}$, then $x=m_1 e_1+m_2 e_2 +\cdots+m_n e_n$ where the $e_i$ are the elements of $R^{n}$ with $1$ in position $i$ and $0$ elsewhere. Therefore $mR^{n}=m^{n}$. Define a map $\pi_n:R^{n}\to (R/m)^{n}$ that reduces each component mod $m$. This is an $R$-module homomorphism whose kernel is $m^{n}$.

Problem 2 continued

  1. The composite map $\pi_k\circ f: R^{n}\to R^{k}/mR^{k}$ has $mR^{n}$ in its kernel.
  2. Suppose that $\pi_k\circ f(x)=0$. Then $f(x)$ belongs to $mR^{k}$, so write $f(x)=m_1 y_1+\cdot+m_j y_j$ with $y_j\in R^{k}$. Since $f$ is surjective, each $y_j=f(x_j)$ and therefore $f(x)=m_1 f(x_1)+\cdots+m_j f(x_j)=f(b)$ where $b\in mR^{n}$. Since $f$ is injective, we see that $x\in mR^{n}$. Thus $\pi_k\circ f$ is surjective and the kernel of $\pi_k\circ f$ is $mR^{n}$.
  3. By the isomorphism theorem, there is a quotient map $\overline{\pi_k\circ f}:R^{n}/mR^{n}\to R^{k}/mR^{k}$ that is an isomorphism. It is also $R/m$ linear – it’s $R$-linear because it’s an $R$-module map, and $m$ acts like zero. Therefore it is an isomorphism of vector spaces between $(R/m)^{n}$ and $(R/m)^{k}$ and therefore $n=k$.

Note: This is false if $R$ is not commutative. Problem 27 constructs an example.

Problem 6

Prove that if $M$ is a finitely generated $R$-module generated by $n$ elements, then any quotient of $M$ can be generated by at most $n$ elements. In particular, quotients of cyclic modules are cyclic.

Problems 16 and 17: Decomposition Theorem

These two problems establish the decomposition theorem for modules. We assume that $R$ is commutative with $1$. Let $M$ be a module and let $I$ and $J$ be ideals of $R$.

  • Prove that the map $M\to M/IM \times M/JM$ defined by $m\mapsto (m+IM, m+JM)$ is an $R$ module homomorphism with kernel $IM\cap JM$.
  • Suppose that $I+J=R$. Prove that the map in the previous part is surjective and its kernel is $IJM$ so that \(M/IJM = M/IM\times M/JM\) in this case.

Decomposition theorem continued

This problem is proved in a manner very similar to the situation for rings. The second part is the interesting one. For surjectivity, given $(m_1+IM,m_2+JM)\in M/IM\times M/JM$, we want to find $m$ so that $m-m_1\in IM$ and $m-m_2\in JM.$ Write $1=i+j$ with $i,j\in R$. Then $m=im+jm$. Let $m_1=jm$ and $m_2=im$. Then $m-m_1=m_2\in iM$ and $m-m_2=m_1\in JM.$

For the second part, suppose $m\in IM\cap JM$. Write $m=im+jm$. Since $m\in JM$, $im\in IJM$, and since $m\in IM$ and $R$ is commutative, $jm\in IJM.$ Therefore $m\in IJM$.

This can be extended by induction to families of ideals $I_1,\ldots, I_k$ that are pairwise relatively prime ($I_i+I_j=R$ for any pair.)

Problem 24 (optional)

This problem constructs an infinite direct product of free modules that is not free; in fact it shows that the countable direct product of copies of $\Z$ is not free.