Maximal ideals in rings of polynomials Let $k$ be a field and $D = k[X_1, . . . , X_n]$ the polynomial ring in $n$ variables over $k$.
Show that:
a) Every maximal ideal of $D$ is generated by $n$ elements.
b) If $R$ is ring and $\mathfrak m\subset D=R[X_1,\dots,X_n]$ is maximal ideal such that $\mathfrak m \cap R$ is maximal and generated by $s$ elements, then $\mathfrak m$ is generated by $s + n$ elements.
The days that I am trying to solve. Help me.
 A: The answer to question a) can be found as Corollary 12.17 in these (Commutative Algebra) notes.  The proof is left as an exercise, but the proof of it is just collecting together the previous results in the section.
(As Patrick DaSilva has mentioned, as written your question b) follows trivially from part a).  I'm guessing it's not what you meant to ask.)
A: We induct on $n$, the result $n=1$ being clear since $k[X_1]$ is a PID. Now consider the inclusion
$$k[X_1,\ldots,X_{n-1}] \subseteq K[X_1,\ldots,X_{n-1}][X_n].$$
Choose a maximal ideal $\mathfrak{m}$ in the ring on the right. Then by Zariski's lemma we have that $k[X_1,\ldots,X_n]/\mathfrak{m}$ is a finite algebraic extension of $k$. Since we have the inclusion
$$k \hookrightarrow k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c \hookrightarrow k[X_1,\ldots,X_n]/\mathfrak{m}$$
it follows that $k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c$ is an integral domain that is also a finite dimensional vector space, thus a field. It will now follow that $\mathfrak{m}^c$ is maximal and the induction hypothesis implies that it is generated by $n-1$ elements.  We now need a Lemma (as suggested by Jeff Tolliver below):


Lemma: $$\left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right) [X_n] \cong k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$$
Proof: $$\begin{eqnarray*} \left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right) [X_n] &\cong& k[X_1,\ldots,X_{n-1}][X_n] \otimes_{k[X_1,\ldots,X_{n-1}]}  k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c \\
&\cong& k[X_1,\ldots,X_n]/\mathfrak{m}^c k[X_1,\ldots,X_n] \hspace{5mm} \end{eqnarray*}$$
    where the first line is using exercise 3.6 of Atiyah - Macdonald, the second line exercise 3.2 of the same book. Explicitly if we trace through the isomorphisms, this sends
    $$\overline{a_0} + \overline{a_1}X_n + \ldots + \overline{a_k}X_n^k \mapsto \overline{a_0 + a_1X_n\ldots + a_kX_n^k}$$
    where the bar on the left is the residue class mod $\mathfrak{m}^c$ and on the right mod $\mathfrak{m}^ck[X_1,\ldots,X_n]$. Thus the lemma is proven.


Back to the problem. It is clear that $\mathfrak{m}$ corresponds to a maximal ideal $\overline{\mathfrak{m}}$ in $k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$ and by the lemma to some maximal ideal in $\left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right)[X_n]$. Since this is the polynomial ring over a field, this maximal ideal is generated by some $P = \overline{a_0} + \ldots + \overline{a_k}X_n^k$. Then in $k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$, the element corresponding to $P$ is $\overline{a_0 + \ldots + a_kX_n^k}$ and this we know generates $\overline{\mathfrak{m}}$. If we write $$f = a_0 + \ldots + a_kX_n^k$$ it will now follow by the induction hypothesis that $\mathfrak{m}$ is generated by $f$ and the $n-1$ elements of $\mathfrak{m}^c$. This completes the proof of the problem.
