Can we always choose the generators of an ideal of a Noetherian ring to be homogeneous? 
Let $R$ be a $k$-subalgebra of $S=k[x_1,x_2,\dots,x_n]$. Let $m\subseteq R$ be the ideal generated by the homogeneous elements of $R$ of positive degree. As $S$ is Noetherian, the ideal $mS$ has a finite set of generators, which may be chosen to be homogeneous elements of $m$. 

Why can we choose homogeneous elements of $m$ to be the generators? 
 A: This is true because $m$ is generated by homogeneous elements, and $mS$ is generated by $m$. Any finite generating set of $mS$ will consist of elements that are linear combinations of elements of $m$ with coefficients in $S$. If the generating set has elements not in $m$, then they'll be of the form
$$\sum{s_im_i}$$
Instead of using this element we can replace it with the finitely many elements $m_1,m_2,\ldots$ and still have a generating set. Similarly, since $m$ is generated by homogeneous elements, if these elements of $m$ are not homogenous we can replace them with homogeneous elements of which they are a linear combination.
A: Elements in $mS$ are $S$-linear combinations of elements in $m$. Take generators of $mS$ and split up each linear combination to obtain generators $m_i \in m$ of $mS$. By definition of $m$, each $m_i$ is an $R$-linear combination of homogeneous elements of $R$. Splitting these up again, gives the desired generators.
A: This is also very confusing to me.  To expand on the answers given, and I'm reading the same book...
$\hat{m} \subset R$ is defined to be the ideal generated by all homogenous polynomials $m$ of varying degree $d \gt 0$.  Say $\hat{m} = (m_i : i \in I)$.  
$\hat{m}S$ is an ideal of $S$ since clearly any $(s_i : i \in J) \subset S$ is an ideal in general.  By definition of $S$, namely that its a polynomial ring over a Noetherian ring, $S$ is Noetherian and so $\hat{m}S = (s_1, \cdots, s_n)$.  Beforehand, we already have that $\hat{m}S = (m_i : i\in I)_S$ so each $s_j = \sum_i m_{i_j} s_i'$. Take $M = \{ $ all $m_{i_j}$ involved $\}$.  Clearly  $(M) = \hat{m}S$.
Now, in a similar manner in which Noetherianity $\equiv$ finitely generated ideals is proved, take the $m_{i_j} \in M$ and form: $(m_{i_{j_1}}) \subset (m_{i_{j_2}}, m_{i_{j_1}}) \subset \cdots = (M) = (f_1, \dots, f_s)$ after relabeling.
Therefore, $\hat{m}S = (f_1, \dots, f_s)_S$ where $f_i$ are homogeneous polynomials in $\hat{m}$.
