I was trying the following: If F is a free abelian group finitely generated by n elements then rank of F is less than or equal to n.

I think we cannot proceed like in vector spaces because here the coefficients are integers and not coming from a field. Then how to do?

  • 1
    $\begingroup$ What is your definition of "rank of (free) abelian group"? For me it is precisely what you give: the cardinality of a minimal set of generators, or the cardinality of any set of free generators. $\endgroup$ – Timbuc Mar 15 '15 at 15:50
  • $\begingroup$ @Timbuc Sir for me it is the cardinality of the basis. $\endgroup$ – akansha Mar 15 '15 at 15:53
  • 1
    $\begingroup$ @akansha Do you have that the cardinality of any spanning set is greater than or equal to the cardinality of any basis? $\endgroup$ – Joe Johnson 126 Mar 15 '15 at 15:55
  • $\begingroup$ @JoeJohnson126 Sir like for the set of integers. $\endgroup$ – akansha Mar 15 '15 at 15:57
  • $\begingroup$ @akansha I'm not sure what you mean. What I mean is, if $\{x_1,\ldots,x_k\}$ is a spanning set of a free abelian group, then any basis has to have less than or equal to $k$ elements. $\endgroup$ – Joe Johnson 126 Mar 15 '15 at 15:59

Write $F \cong \mathbb Z^r$. Consider the homomorphism $\pi: F \cong \mathbb Z^r \to (\mathbb Z/2\mathbb Z)^r = \mathbb F_2^r$ given by the projection $\mathbb Z\to \mathbb Z/2\mathbb Z$ on each component. If $F$ is generated by $n$ elements $x_1,\dots,x_n$, then their images $\pi(x_1),\dots,\pi(x_n)$ span the $r$-dimensional vector space $\mathbb F_2^r$, hence we must have $n\geq r$.

(Obviously there is nothing special about the choice of prime 2 in this proof. Any other prime would also work.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.