# about simple modules over $\mathbb{Q}[x,y,z]$

Let $R=\mathbb{Q}[x,y,z]$, then every simple $R$-module $M$ is finite dimensional over $\mathbb{Q}$.

Had this been over $\mathbb{C}$ (complex field), it would have been rather easy. I have tried to use a theorem which says simple modules over $R$ is isomorphic to $R/I$, where I is a maximal regular ideal. But I don't understand regular ideals all that well. (For example, $Q[x,y]/(xy-1) \simeq Q(y)$, but $Q(y)$ is not finite dimensional over $Q$.)

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Out of curiosity, how are you solving the problem with $\mathbb{Q}$ replaced by $\mathbb{C}$? Are you using Hilbert's Nullstellensatz? –  Pete L. Clark Jan 30 '11 at 1:13
if $R$ is a domain which is not a field, every nonzero ideal is regular (i.e., contains a non-zero divisor), so "maximal regular" = "maximal". The ideal $(xy-1)$ is not a maximal ideal of $\mathbb{Q}[x,y]$, and the quotient by that ideal is isomorphic neither to $\mathbb{Q}(t)$ nor $\mathbb{Q}[t]$ but to $\mathbb{Q}[t,t^{-1}]$. (You seem a bit confused...) –  Pete L. Clark Jan 30 '11 at 1:18
all maximal ideals of R= C[x,y,z] look like I=(x-a,y-b,z-c) by Nullstellensatz so then it looks like $\bar x$ is algebraic and etc. i was wrong about (xy-1) being maximal. once i made an homomorphism sending x to t and y to t^{-1} i assumed i had a field but that's wrong. thanks for what you have written. –  Heidi Jan 30 '11 at 6:35

## 1 Answer

This follows easily from a result known as "Zariski's Lemma": if $k$ is a field, $k'$ an extension which is a finitely generated $k$-algebra, then $k'$ is a finite extension of $k$. So for any maximal ideal $I \subset k[x_1, \dots, x_n]$, the ring $k[x_1, \dots, x_n]/I$ is a finite field extension of $k$. For a short proof of this due to McCare, see section 7.5 of http://people.fas.harvard.edu/~amathew/CRing.pdf (in particular the exercise at the end for a really quick one, not using the machinery of the Noether normalization theorem).

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By the way, I don't know what a regular ideal is. But $R/I$ is simple iff $I$ is maximal, and every simple module is isomorphic to one of this form. –  Akhil Mathew Jan 30 '11 at 0:32
What you said is correct. What I came to realize is that this problem is stated incorrectly. Just because a field is a finite extension over another field, it doesn't mean that it's 'finite dimensional' over the other field as a vector space. –  Heidi Jan 30 '11 at 1:06
@Heidi: Dear Heidi, I don't understand. A field that is a "finite extension" of another field is by definition finite-dimensional over the smaller field. A field that is a finitely generated field extension is not f.d. (e.g. a field $k(t)/k$), but the point of Zariski's lemma is that if the overfield is finitely generated as an algebra (not just as a field!), then it is furthermore finite algebraic. –  Akhil Mathew Jan 30 '11 at 1:12
i was trying to say Q(x) is finitely generated as an algebra over Q (unless i'm mistaken). but Q(x) over Q as a vector space contains all basis x, x^2,...., x^n.. and hence is infinite dimensional. since the question said R-module, i automatically thought of M as a vector space over Q. –  Heidi Jan 30 '11 at 6:44
and i see the subtleties you've pointed out. thanks a lot! –  Heidi Jan 30 '11 at 6:49