Let ${\displaystyle F}$be a field, ${\displaystyle R=F[x_{1},\dots ,x_{n}]}$(note: finitely many variables), and ${\displaystyle M}$be a maximal ideal of ${\displaystyle R}$. So ${\displaystyle R/M}$is a field extension of ${\displaystyle F}$. Is it necessarily an algebraic extension? Antendren (talk) 23:23, 22 February 2025 (UTC)[reply]

Yes. Let ${\displaystyle x}$be a generator of a transcendental extension of ${\displaystyle F}$in ${\displaystyle R}$. Then ${\displaystyle x}$is a polynomial in the ${\displaystyle x_{i}}$and the ideal generated by ${\displaystyle M}$and ${\displaystyle x}$is proper, i.e., ${\displaystyle M}$is not maximal, contradiction. Tito Omburo (talk) 00:20, 23 February 2025 (UTC)[reply]

Why is it proper? And how are you using that there are only finitely many variables, since it's not true otherwise?--Antendren (talk) 00:31, 23 February 2025 (UTC)[reply]

The ideal is proper because it does not contain F (x is transcendental over F). Are you certain it's not true for infinitely many variables? Tito Omburo (talk) 00:45, 23 February 2025 (UTC)[reply]

I don't see why x being transcendental makes the ideal proper. Could you give the details?

Yes, it's not true for infinitely many variables. Let ${\displaystyle F=\mathbb {Q} }$, let ${\displaystyle K}$be an extension by a single transcendental element, and let ${\displaystyle \{\alpha _{i}:i\in \mathbb {N} \}}$list the elements of K. Define a homomorphism from ${\displaystyle F[x_{1},x_{2},\dots ]}$to ${\displaystyle K}$by ${\displaystyle x_{i}\mapsto \alpha _{i}}$, and let M be the kernel.

Note that in this case, some ${\displaystyle x_{i}}$corresponds to a transcendental element, and some ${\displaystyle x_{j}}$corresponds to its inverse, so ${\displaystyle M}$contains ${\displaystyle x_{i}x_{j}-1}$, meaning that ${\displaystyle (M,x_{i})}$isn't proper.Antendren (talk) 00:59, 23 February 2025 (UTC)[reply]

To ask this question another way: Suppose ${\displaystyle K}$is a transcendental extension of ${\displaystyle F}$. As a vector space, ${\displaystyle K}$is not finitely generated over ${\displaystyle F}$. As a field, it might be (${\displaystyle {\mathbb {Q} }(\pi )}$over ${\displaystyle \mathbb {Q} }$). What about as a ring?--Antendren (talk) 11:09, 24 February 2025 (UTC)[reply]

For anyone curious, it turns out the answer was Zariski's lemma.--Antendren (talk) 04:12, 5 March 2025 (UTC)[reply]