TITLE: Homogeneous elements and graded algebra - any misconception?
QUESTION [5 upvotes]: Elements of any factor $A_n$ of the decomposition are known as
  homogeneous elements of degree $n$. An ideal or other subset
  $\mathfrak{a} ⊂ A$ is homogeneous if every element $a ∈ \mathfrak{a}$
  is the sum of homogeneous elements that belong to $\mathfrak{a}.$ For
  a given $a$ these homogeneous elements are uniquely defined and are
  called the homogeneous parts of $a$. Equivalently, an ideal is
  homogeneous if for each $a$ in the ideal, when $a=a_1+a_2+...+a_n$
  with all $a_i$ homogeneous elements, then all the $a_i$ are in the
  ideal.

I am greatly confused. So, by saying  "An ideal or other subset $\mathfrak{a} ⊂ A$ is homogeneous if every element $a ∈ \mathfrak{a}$ is the sum of homogeneous elements that belong to $\mathfrak{a}.$" and "Elements of any factor $A_n$ of the decomposition are known as homogeneous elements of degree $n$.", 
what exactly is homogeneous elements? So we call elements of the same factor of the decomposition as being homogeneous elements of same degree? And by saying "$a_i$ homogeneous elements", is this saying that $a_i$ and $a_j$ where $i \neq j$ cannot be homogeneous elements of the same degree?
Can anyone correct misconception of homogeneous elements?

REPLY [11 votes]: Maybe an example would help.  The ring $A = k[x, y]$ is graded by total degree, so $A_n$ is the linear span of monomials $x^ay^b$ where $a + b = n$.
The element
$$a = (y + 1)(x - y^2) = x + xy - y^2 - y^3$$
is the sum of homogeneous parts
$$a_1 = x$$
$$a_2 = xy - y^2$$
$$a_3 = - y^3$$
All other $a_i$ are zero.  So we're just splitting apart the degree $1$, degree $2$, and degree $3$ stuff from $a$.  By definition each $a_i$ has degree $i$, so the elements $a_i \in A_i$ are homogeneous but $a$ itself is not homogeneous because it is not contained in a single $A_i$, it contains monomials of different total degree.
Now consider the ideal $I = (x, y^2)$.  This ideal is homogeneous (if you haven't done so yet you will soon prove that an ideal is homogeneous if and only if it can be generated by homogeneous elements).  Note that it contains the element $a$ above because $x - y^2 \in I$.  Note that each homogeneous part $a_i$ of $a$ is also contained in $I$.
Next consider the ideal $J = (y + 1)$ which also contains $a$.  This ideal is not homogeneous.  Indeed, it contains $a$ but it does not contain the homogeneous part $a_1$ of $a$.
Edit: One last comment.  When I say "an ideal is homogeneous if and only if it can be generated by homogeneous elements" I specifically chose the word can instead of the word is.  I can also write $I = (x-y^2, y^2)$ and now $I$ is not generated by homogeneous elements, but it can be generated by homogeneous elements if I chose different generators.  On the other hand there is no choice of homogeneous elements that will generate $J$.  That's not always obvious, hence we needed to observe that $a \in J$ while $a_1 \notin J$ to conclude that $J$ is not homogeneous.