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This may be a trivial question. We say an ideal $I$ in a ring $R$ is $k$-generated iff $I$ is generated by at most $k$ elements of $R$. Let $F$ be a field. Is it true that every ideal in $F[x_1,x_2,....,x_n]$ is $n-$generated. (This is true when $n=1$, because $F[x_1]$ is a PID)

Second question: Is it true that every ideal in $F[x_1,x_2,x_3,...]$ is generated by a countable set of elements of $F[x_1,x_2,x_3,...]$ ?

Thank you

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As F is Noetherian, K = F[x1,..xn] is also, so every ideal in K is f.g., and <x1, x2...xn,2 > ideal generated in C[x1,..xn], this is generated by n+1 elements, am I wrong some where? – Ram Jan 22 '13 at 10:11
@Ram: if $2 = 0$ then that ideal is generated by $n$ elements; if $2 \neq 0$ then it is invertible, so that ideal is the unit ideal and is generated by one element. – Qiaochu Yuan Jan 22 '13 at 10:14
@YACP I am self studying abstract algebra (still a beginner). Thus, many of the terms that were mentioned in my previous question (like Krull's dimension, Hillbert Samuel polynomial...) I never heard of. As a result I didnt really digest Cohen's theorem. – Amr Jan 22 '13 at 20:01
@YACP The reason why I asked this question is the following: I wanted to find an analouge to the division algorithm in $F[x,y]$. I conjectured that for any polynomials $f(x,y),p(x,y),q(x,y)$ (p,q relatively prime) there exists polynomial $g(x,y)$ such that $deg(g)< max\{deg(p),deg(q)\}$ and $f-g\in(p,q)$. – Amr Jan 22 '13 at 20:07
@YCAP Finally, I was hoping that I could use this to show that every ideal $I$ in $F[x,y]$ is 2-generated – Amr Jan 22 '13 at 20:17
up vote 6 down vote accepted

The answer to the first question is no. For example, the ideal $(x^2, xy, y^2)$ in $F[x, y]$ cannot be generated by $2$ elements. To see this, note that any set of generators must linearly span the subspace of homogeneous quadratic polynomials, which has dimension $3$.

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I see. Any thoughts about the second question – Amr Jan 22 '13 at 10:16
Is there a function $f(n)$ such that every ideal in $F[x_1,x_2,...,x_n]$ is $f(n)$ generated ? – Amr Jan 22 '13 at 10:18
Qiaochu's example can be easily generalized: consider the ideal generated by all monomials of degree $d$. Their number is unbounded as $d$ grows. – Andrea Mori Jan 22 '13 at 10:21
I am not sure what the answer for radical ideals is though. – Qiaochu Yuan Jan 22 '13 at 22:07

Since Qiaochu has answered your first question, I'll answer the second: yes, every ideal $I\subset F[x_1,x_2,x_3,...]$ is generated by a countable set of elements of $F[x_1,x_2,x_3,...]$.

Indeed, let $G_n\subset I_n$ be a finite set of generators for the ideal $I_n=I\cap F[x_1,x_2,x_3,...,x_n]$ of the noetherian ring $F[x_1,x_2,x_3,...,x_n]$.
The union $G=\bigcup_n G_n$ is then the required denumerable set generating the ideal $I$.
The reason is simply that every polynomial $P\in I$ actually involves only finitely many variables $x_1,...,x_r$ so that $P\in F[x_1,x_2,x_3,...,x_r]$ for some $r$ and thus, since $P\in I_r$, one can write $P=\sum g_i\cdot f_i$ for some $g_i\in G_r\subset G$ and $f_i\in F[x_1,x_2,x_3,...,x_r]$.
This proves that $G$ generates $I$.

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Yes. I also figured another solution. I will put it shortly (+1) – Amr Jan 22 '13 at 13:02

An ideal is finitely generated if it is finitely generated as $R$-modulo. So your first statement is wrong because there are at least the base-ring. For your second question note that Hilbert basis theorem say that if $R$ is a noetherian ring also $R[x]$ is a notherian ring. But you know that notherian rings are finitely generated as $R$-moduli.

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Unfortunately, I don't know Hilbert basis theorem and I still didnt study moduli – Amr Jan 22 '13 at 10:27
I can't make sense of your second and fourth sentences. In the third sentence the theorem you invoke is not the basis theorem, and anyway has no relevance to Amr's second question. – Georges Elencwajg Jan 22 '13 at 12:56
$R[x_1,x_2,...]$ is not Noetherian, take the ascending chain of ideals $ (x_1) \le (x_1, x_2) \le (x_1,x_2,x_3)...$ has no maximal element. – Ram Jan 22 '13 at 13:00

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