ring seen as vectorspace over subfield

fredbel6

ring seen as vectorspace over subfield

**Posted:** Thu Dec 16, 2004 6:17 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

hi,

it looks like there's some condition missing; i guess R is a finite-dimensional K-vector space(else, consider e.g. the direct product of an infinite number of copies of $\mathbb{Q}$ : the subset of those elements with all coordinates equal gives a subring which is isomorphic to $\mathbb{Q}$ and all the ideals consisting of those elements which are at the i-th place zero are maximal).
i'm not quite ready with the second thing; it must be for the reason i'm totally blocked at the moment.

let $\mathcal P$ be any prime ideal. let $R'=R/{\mathcal P}$ . since $\mathcal P$ does not contain any element of $K$ , $K\to R/{\mathcal P}$ is injective. furthermore, $R/P$ is still finite-dimensional(in fact the old basis now generates $R/P$ ). therefore it suffices to prove that any finite-dimensional K-vector space $R$ which is an integral domain is in fact a field. let $x\in R$ ; consider $K[x]$ . it is indeed finitely-generated since $R$ is(remember that K-modules are vector spaces); therefore there is a minimal polynomial $P(x)=a_nx^n+...+a_0$ for $x$ over $K$ . since $R$ has no zero divisors, $a_0\neq 0$ , therefore $x^{-1}=-a_0^{-1}(a_nx^{n-1}+...+a_1)$ .

the next thing trivially follows since the intersection of all maximal ideals is the intersection of all prime ideals which in turn is the set of all nilpotent elements which is 0 by the claim.

for the last thing, i guess we may again assume there are no nilpotent elements; otherwise consider $R=\mathbb{R}[x]/x^2$ . it has the subring $\mathbb{R}$ which is a field and it is a finite-dimensional $\mathbb{R}$ -vector space. on the other hand, if it is product of two rings, then there are elements $e_1, e_2$ s.t. $e_1+e_2=1$ , $e_1e_2=0$ , $e_1^2=e_1$ , $e_2^2=e_2$ . but the only zero divisors are multiples of $x$ , so $e_1=a_1x$ and $e_2=a_2x$ , contradicting $e_1+e_2=1$ .
if it has no nilpotent elements, let $m_1,...,m_k$ be the maximal ideals. then, by the chinese remainder theorem, $R=R/m_1\times ... R/m_k$ ; since the ideals are all maximal, the factors on the right are all fields.

**Posted:** Thu Dec 16, 2004 9:44 pm

fredbel6

#3

indead, R is a finite dimensional vector space over K, i forgot that

but there are many things i don't get

for starters : why is the intersection of all prime ideals in a ring the
set of nilpotent elements?

fred

**Posted:** Thu Dec 16, 2004 10:43 pm

grobber

#4

That is well-known, but not at all obvious.

If $x$ is nilpotent, then you can easily see that it belongs to every prime ideal, because $x^n=0$ belongs to every prime ideal.

Conversely, consider the set of ideals which do not contain $x^n$ for any $n>0$ . Show that this set, partially ordered by inclusion, is inductive, so it has a maximal element (Zorn's Lemma). It's not at all difficult to show that this maximal element is a prime ideal.

**Posted:** Fri Dec 17, 2004 2:11 am

fredbel6

#5

hi

thx for the many remarks
so i have to use zorn's lemma ( i kind of dislike doing axiom of choice
stuff) Sad

however, does anyone have an idea for the other problems

i understand that proving finiteness of maximal ideals is the same
as prime ideals?

is that easier?

**Posted:** Fri Dec 17, 2004 8:32 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#6

i don't think so since in general, there are much more prime ideals than maximal ideals, so in order to prove that one of these sets is finite, it should be easier to prove that there are only a finite number of maximal ideals(one even has more conditions).

btw, one often has to use zorn's lemma in algebra; i disliked it first, but after having seen how nicely it may be used, i now love to use it Very Happy

**Posted:** Sat Dec 18, 2004 2:36 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#7

btw, why do you say "other problems" in the plural? they're only the problem about the finite number of the miaxmal ideals left.

**Posted:** Sat Dec 18, 2004 2:37 pm

fredbel6

#8

sorry second time i make it unnecessarily difficult for you people here
indead, the ring is a finite dimensional space over K, and there are no
nilpotent elements in the : prove that it is isomorphic with a direct
product of fields-question

so indead the big question is why are there only a finite amount of maximal ideals
i realize that the ideals are all finite dimensional spaces over K
but what else can you say?

btw : peter, my problem with the axiom of choice or zorns lemma is
that (in my course) it pops up when we want to prove that all
systems of equations in n variabeles
( that generate an ideal not the entire polynomial ring because that surely would lead to a contradiction) is solvable in a field extension
u need the fact that every (not the full ring) ideal is contained in a
maximal ideal, later , it turned out to be possible to do with hilberts basis theorem
isn't the axiom only supposed to be used when u are really sure
u can't do it without, because otherwise you make it depending on a
rejectable axiom unnecessarily

**Posted:** Sat Dec 18, 2004 3:39 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#9

yes, the use of zorn's lemma and the axiom of choice is often tried to put in very low measures and one always tries to find a proof without it or else tries to prove that it's equivalent to the axiom of choice.
but however, if you really need to use it in the very foundations of some theory(and commutative algebra might be counted to those...for example the very fundamental theorem that we always got a maximal ideal) then it is very likely that any stronger sentences need to go down to that theorem at some point. and if so, you may use zorn's lemma in that proof as much as you want since you need it anway for the citation of that very basic theorem.

furthermore, strictly speaking every moment you have a set and you want to show it's empty by taking one element and showing that it doesn't exist, you have to use the axiom of choice. and that appears quite frequently.

however, that's not the point.

indeed, the last question requires that the number of maximal ideals is finite. i still don't quite got a proof for this result. but one can show that every member of a maximal ideal is a zero divisor. in fact, consider a basis $a_1,...,a_n$ for R over K and let $b\in m$ . then if $ba_1,...,ba_n$ were linearly independet they were a basis; but then since $ba_1,...,ba_n\in m$ , $m$ would contain whole R. therefore they are linearly dependent and we get $(x_1a_1+...+x_na_n)b=0$ where $x_i\neq 0$ for some $i$ ; therefore $x_1a_1+...+x_na_n\neq 0$ since $a_1,...,a_n$ form a basis therefore are linearly independent.

i do not quite see how this helps us. by the way this gives another proof for the thing that every prime ideal is maximal; in fact every unit is in some maximal ideal and therefore is a zero divisor by the above. therefore there are only units and zero divisors in R. now, in $R/{\mathcal P}$ (notation as above) there are only units, therefore this is a field.

**Posted:** Sat Dec 18, 2004 4:24 pm

fredbel6

#10

hi
i think i have proof of the first, but could someone check it plz

every quotient ring of R is also a finite dimensional space over K

suppose there are infinitely many maximal ideals

take n of them
all of them are comaximal , so by chinese remainder theorem:

R/I1*.....*R/In =R/(I1*..*In)
the left part is a K vectorspace with as dimension the product of all
dimensions of quotientrings, which is at least 2^n
so i could make quotient rings with dimension as big as i want, which is impossible

what do you think? i'm posting it not only to have it checked, but also because u might like it as several people have helped me
(I got the idea when giving up on it and trying the next problems)

also ; the full problem also had these problems

take K=C (complex numbers), R a ring containing C, R has no
nilpotent elements and is finite dimensional over C

proof that R is isomorphic with a coordinate ring over an algebraic variety in C (onedimensional affine space) and that such variety needs to be finite

i think i can do them but i'm open to all comments about it

**Posted:** Sat Dec 18, 2004 9:44 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#11

argh...why i didn't get this Sad

anyway, it is actually a very beautiful proof.

R is finitely generated over $\mathbb{C}$ ; let $x_1,...,x_n$ be generators and let $I$ be the kernel of $\mathbb{C}[x_1,..,x_n]->R$ . then $R=\mathbb{C}[x_1,..,x_n]/I$ . we know that $r(I)=I$ since $R$ has no nilpotent elements. therefore the variety defined by $I$ (the points, where $I$ is 0), has the associated ideal $r(I)=I$ (hilbert's nullstellensatz), thus the coordinate ring is isomorphic to $R$ . i guess a variety is finite if it is finite-dimensional(am i wrong?), but this is clear since $n$ is finite.

Peter

**Posted:** Wed Dec 22, 2004 9:40 pm

fredbel6

#12

your proof looks a lot like the professor gave us

however the argument for finiteness of the variety was that there are only finite amount of maximal ideals, and hilberts nullstellensatz makes correspondende between zeroes of ideals and maximal ideals containing them

**Posted:** Sun Dec 26, 2004 11:26 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#13

? i guess there is more than just a finite number of maximal ideals since maximal ideals correspond to the points of that variety; thus a finite variety would consist of just a finite number of points! this sure is not always the case! so, what's the actual definition of a finite variety?

**Posted:** Mon Dec 27, 2004 9:38 pm

fredbel6

#14

wait a minute peter
that indead doesn't make much sense!
Sad

because it was the so called coordinate ring with which the ring R we began with was isomorphic?
for this statement to be true, C[x1,...,xn]would have only finitely many
maximal ideals

so i'm open to all help
i'll make inquiries about this matter Blush

**Posted:** Tue Dec 28, 2004 12:03 am

fredbel6

#15

i think i found it
the nullstellensatz makes correspondence between maximal ideals containing the ideal and the zeroes of the ideal

because of the correspondence theorem :
every maximal ideal in C[x1,...,xn] containing I, corresponds to
a maximal ideal in C[x1,...,xn] /I

and C[x1,...,xn]/I is isomorphic with the ring we began with, --->
has finite amount of maximal ideals---> there is finite amount of maximal ideals in C[x1,...,xn] containing I---> I has a finite variety

does this make sense? Razz

**Posted:** Tue Dec 28, 2004 12:08 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

#16

hmm...

now i think i see the point were we both talked about something different. i thought R is finitely generated as a $\mathbb{C}$ -algebra; in this case, R is just a finite-dimensional variety, but not finite; but i guess you were given the condition that R is finitely generated as a $\mathbb{C}$ -module(= $\mathbb{C}$ -vector space); in this case, of course, by the above, the variety is finite.

**Posted:** Wed Dec 29, 2004 2:38 pm