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Let $R$ be a discrete valuation ring and let $A$ be a $R$-algebra of $R[X]$ and let $A'$ be a $R$-subalgebra of $R[X]$.

It has a property that homomorphic image of a finitely generated algebra also has this property however homomorphic image of a finitely generated subalgebra may not be again finitely generated.

But I don't have an example for subalgebra.

Why is so ? what is the problem with homomorphic image of subalgebra to hold the property.

I need an example to see this.

For example, consider the polynomial ring $R[X_1,X_2, \cdots,X_n]$ of the field $K[X_1,X_2, \cdots, X_n]$.

Any finitely generated $R$-algebra of $R[X_1,X_2, \cdots,X_n]$ is isomorphic to $R[X_1,X_2, \cdots,X_n]/I$, where $I \subset R[X_1,X_2, \cdots,X_n]$ is an ideal.

How to define a map so that homomorphic image of a finitely generated subalgebra is not finitely generated subalgebra ?

We see that $P=R[X_1,X_2]$ is a finitely generated subalgebra of $R[X_1,X_2, \cdots,X_n]$ and define another subalgebra $Q=R[X_1,X_1^2X_2, X_1^3X_2, X_1^4X_2, \cdots]$, which is not finitely generated.

How to define a homomorphism between $P$ and $Q$ ?

You can give other example as well.

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    $\begingroup$ A homomorphic image of a finitely generated thing is always finitely generated, with generators given by the image of the generators of the original. $\endgroup$ – Qiaochu Yuan Sep 25 '20 at 19:26
  • $\begingroup$ @QiaochuYuan, I think it is not true as mentioned in $\text{properties section}$ in the link here. Can you please check it and confirm? $\endgroup$ – Masmath Sep 26 '20 at 3:34
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    $\begingroup$ Wikipedia says exactly what I said: a homomorphic image of a finitely generated thing is finitely generated. It then says something else, which is that a subalgebra of a finitely generated algebra need not be finitely generated, but that’s a different statement. $\endgroup$ – Qiaochu Yuan Sep 26 '20 at 3:37
  • $\begingroup$ @QiaochuYuan, oh I see. But it was looking like same $linked$ statement which confused me. $\endgroup$ – Masmath Sep 26 '20 at 3:54
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As Qiaochu said in the comments, a homomorphic image of a finitely generated object is always finitely generated. If $A$ is finitely generated (say, as a $k$-algebra, where $k$ is some commutative ring) and $\phi : A\to B$ is a surjective $k$-algebra homomorphism, we may write $A = k[x_1,\dots, x_n]/I$ for some ideal $I$. But since $A\to B$ is a surjection, $B\cong A/\ker\phi\cong k[x_1,\dots, x_n]/(I,\ker\phi).$ So $B$ is also finitely generated. Explicitly, if $A$ is generated by $a_1,\dots, a_n,$ then $B$ is generated by $\phi(a_1),\dots, \phi(a_n).$

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