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Let $M$ be a finitely generated $R$-Module and let $N$ be a free module. Let $f: M \rightarrow N $ be a surjective homomorphism. Show that $N$ has a finite rank and that kernel of $f$ is finitely generated.

I have used the finite rank and other properties given to show that Kernel of $f$ is finitely generated but I don't know how to show that $N$ has a finite rank. Thanks for your help.

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    What is a ring for you? Unity? Commutative? – 2017-02-16
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    If it is of infinite rank, then you have an infinite set of linear independent elements whose preimages are also linearly independent, thus contradicting the finiteness property of M. – 2017-02-16
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    Yes, our ring is commutative and has a unity. – 2017-02-16
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    I do not know how you have proven the statement about the kernel, but note that in this case $M = \operatorname{ker}(f) \oplus N$ holds, hence $\operatorname{ker}(f) \cong M/N$ is a homomorphic image of $M$ and thus finitely generated. – 2017-02-16
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    Thank you all. I think I am good. Thanks again – 2017-02-16

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$M$ is finitely generated, so $M/\ker(f)$ is finitely generated, but $M/\ker(f)\cong N$. Of course, a free module is f.g. iff it is of finite rank.

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    Thank you. I got what I wanted. – 2017-02-16