Let be a ring homomorphism. We say that is a finitely-generated as an -algebra if there exists a surjection

We say that is finitely-presented if it is finitely-generated and we can choose to be a finitely-generated ideal.

If there exists a such that is finitely-generated, then for all other , is finitely-generated due to the lemmalemma below. It is the analogue of Schanuel’s lemmaSchanuel’s lemma for algebras.

has a non-finitely-generated kernel.

Let be a ring and be a finitely-presented -algebra. If is any surjection of -algebras, then is a finitely-generated ideal.

Let be a presentation, where is a finitely-generated ideal. Form the following diagram with exact rows:

Because is surjective, we can find a lift of to

which gives us the middle vertical map. By the Snake Lemma, we get a surjection

Since is a finitely-generated -module, it is certainly a finitely-generated -module via the horizontal map

Therefore is a finitely-generated -module. From vertical exact sequence on the left, we get a short exact sequence

and so is a finitely-generated -module by 3.3. of this propositionthis proposition; note here that is a -module, and becomes a -module via the vertical map

But because this map is surjective, is a finitely-generated ideal of .

Consider -algebras and in the following diagram:

Viewing and as just -modules, we can take their tensor product , and there are -linear maps

Since the ring multiplication in and are -bilinear, we get two induced maps out of the tensor products

Tensoring these two maps together and reshuffling the tensor factors,

we get an -bilinear multiplication on . This makes into an -algebra. It satisfies the following universal property: for all -algebra maps , there exists a unique -algebra map such that the following diagram commutes

This is the coproduct in the category of -algebras.

Compare this to the coproduct in the category of sets:

Note that the nature of the coproduct for -algebras is quite different.

Taking , we get

Recall that we can take the fibre product as sets. In this case, we get a unique map which fits in the following diagram:

However, and the fibre product are quite different.

since and are both just one-point spaces. But if we instead consider ,

we have

by the Chinese Remainder Theorem. Thus .

However, for example, but does not come from either or .

Let and be -algebras. Then is a -algebra and every -algebra map , where is a -algebra, admits a unique factorisation through , i.e.,

Let be a ring. Let be an -module. We say that is Noetherian if every ascending chain of -submodules

eventually stabilises.

Let be a ring and be an -module.

Consider an exact sequence of -modules

Then:

is Noetherian if and only if every -submodule is finitely-generated.

If is Noetherian over itself, then the following are equivalent:

1.1. and 2.2. are left for the worksheets. We prove 3.3.. 3.1.3.1. 3.23.2 is OK. 3.3.3.3. 3.2.3.2. is by 2.2.. We prove 3.2.3.2. 3.3.3.3. and 3.2.3.2. 3.1.3.1.. By induction on and 1.1.1.1. in the following exact sequence,

we find that is Noetherian for all . Now, if is finitely-generated, then there is a presentation

Since is Noetherian, is Noetherian and is Noetherian. Then by 2.2., is finitely-generated.

is a Noetherian ring if and only if every ideal is finitely-generated.

Ideals of are -submodules of .