# Does a homomorphism send a finite set to a finite set?

Does a homomorphism send a finite set to a finite set ?

I know from the group theory that, instead of a homomorphism, if we have an isomorphism $\phi :G\to\bar G$ and $G=\langle g\rangle$ then $\bar G=\langle\phi(g)\rangle$.

Is this still valid in case of a ring homomorphism ?

I mean does the number of generators of an ideal stay finite under a ring homomorphism ?

(I have to prove that if $R$ is noetherian then so is $S^{-1}R$. I could show that there is $1-1$ correspondence between the ideals of them, but not finiteness of the generators. )

• In general, any application sends a finite set onto a finite set. Since an application sends an element of the domain to exactly one element in the codomain, then the range has less elements than the domain. – Taladris Feb 21 '15 at 14:15
• there is not 1−1 correspondence between the ideals of them – user 1 Feb 21 '15 at 14:18
• Yes if $\phi:R\to S$ is a ring isomorphism and $I=(a_1,\cdots,a_d)$ then $\phi(I)=(\phi(a_1),\cdots,\phi(a_d))$. In general if your ring homomorphism is not onto then the image of an ideal needn't be an ideal however. – whacka Feb 21 '15 at 14:19
• @user 1 sorry I meant any ideal in $S^{-1}R$ is the image of some ideal in $R$ – implicit lee Feb 21 '15 at 14:19
• Note: My comment above answers to the titular question. – Taladris Feb 21 '15 at 14:21

For your specific question of the noetherianness of $S^{-1}R$, the correspondance between ideal of $S^{-1}R$ and ideals of $R$ is simply this:
Let $i\colon R\rightarrow S^{-1}R$ the canonical homomorphism. If $\mathfrak a$ is an ideal in $S^{-1}R$ and $\mathfrak b=i^{-1}\mathfrak a$, then $\mathfrak a=S^{-1}\mathfrak b$.
Thus, if $R$ is noetherian, $\mathfrak b$ has a finite set of generators $b_1,\dots, b_r$, and $\mathfrak a$ is generated by $\dfrac{b_1}1,\dots,\dfrac{b_r}1$.
• thanks here it is proven for submodules instead of ideals, but there is no difference I think. A stupid question: Does canonical map not mean that the domain is ''bigger'' than the range and the map should be onto ? Here $R$ is a subring of $S^{-1}R$. – implicit lee Feb 21 '15 at 15:02
• Here the canonical map just sends $x$ to $\dfrac x1$. It has no reason to be onto, nor one-to-one either, except if $R$ is an integral domain. – Bernard Feb 21 '15 at 15:11