# homomorphic image of a finitely generated subalgebra may not be again finitely generated

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.

• A homomorphic image of a finitely generated thing is always finitely generated, with generators given by the image of the generators of the original. – Qiaochu Yuan Sep 25 '20 at 19:26
• @QiaochuYuan, I think it is not true as mentioned in $\text{properties section}$ in the link here. Can you please check it and confirm? – Masmath Sep 26 '20 at 3:34
• 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. – Qiaochu Yuan Sep 26 '20 at 3:37
• @QiaochuYuan, oh I see. But it was looking like same $linked$ statement which confused me. – Masmath Sep 26 '20 at 3:54

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