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In Miles Reid's Undergraduate Commutative Algebra he defines a ring $B$ to be finite as an $A$ - algebra if it is finite as an $A$ - module. Now what I don't understand is suppose we look at the polynomial ring $k[x_1,\ldots,x_n]$ where $k$ is a field. Then as a $k$ - algebra it is finitely generated. Is this the same as being a finite $k$ - algebra? For if it is the same this means that $k[x_1,\ldots,x_n]$ is finitely generated as a $k$ - module which is just a $k$ - vector space. However this cannot be possible because the $x_i's$ are not even algebraic over $k$. What am I misunderstanding here?

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It might be a typo...? What you define in your first paragraph is commonly known as a finite $k$-algebra. –  Zhen Lin May 20 '12 at 12:18
    
@ZhenLin Is there a difference between a finite $k$ - algebra and a finitely generated $k$ - algebra? –  user38268 May 20 '12 at 12:19
    
@ZhenLin Indeed I have misread it, he defined it as a finite $k$ - algebra and not a finitely generated $k$ - algebra. It seems I got confused between the two definitions. –  user38268 May 20 '12 at 12:22
    
Yes. It's an annoying quirk of mathematical English, unfortunately. A finite $k$-algebra is finitely generated as a $k$-module, but a finitely-generated $k$-algebra usually is not. –  Zhen Lin May 20 '12 at 12:25
    
@ZhenLin Thanks I tacitly assumed the two were equivalent. –  user38268 May 20 '12 at 12:27

1 Answer 1

up vote 7 down vote accepted

An $A$-algebra $B$ is called finite if $B$ is a finitely generated $A$-module, i.e. there are elements $b_1,\dotsc,b_n \in B$ such that $B=A b_1 + \dotsc + A b_n$. It is called finitely generated / of finite type if $B$ is a finitely generated $A$-algebra, i.e. there are elements $b_1,\dotsc,b_n$ such that $B=A[b_1,\dotsc,b_n]$. Clearly every finite algebra is also a finitely generated one. The converse is not true (consider $B=A[T]$). However, there is the following important connection:

An algebra $A \to B$ is finite iff it is of finite type and integral.

For example, $\mathbb{Z}[\sqrt{2}]$ is of finite type over $\mathbb{Z}$ and integral, thus finite. In fact, $1,\sqrt{2}$ is a basis as a module. You can find the proof of the claim above in every introduction to commutative algebra.

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