A question about polynomial rings 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
 A: 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 $3$. 
A: 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 $I$.
A: 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.
