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.
Best Answer
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 $\phi(a_1),\dots, \phi(a_n).$