Homogeneous elements and graded algebra - any misconception?
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?
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$\begingroup$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.
$\endgroup$ $\begingroup$Yes this text is a bit confusing. Better use the canonical texts on commutative algebra to learn these basics about graded rings and graded ideals.
The underlying abelian group of a graded ring $A$ is the direct sum of certain subgroups $A_n$, where $n \in \mathbb{Z}$. The elements of $A_n$ are called homogeneous of degree $n$. Every element is a unique finite direct sum of homgeneous elements $a=\sum_n a_n$, where $a_n$ is homogeneous of degree $n$ and almost all $a_n=0$. Here, $a_n$ is called the homogenous part of $a$ of degree $n$. It is just the image of $a$ under the canonical projection $A \to A_n \to A$.
An ideal of the underlying ring of a graded ring is called graded if it can be generated by homogeneous elements. Equivalently, the homogeneous components of an element of the ideal also belong to the ideal. A better and more categorical definition: Graded ideals are exactly the kernels of graded ring homomorphisms.
It is not correct that if $a,b$ are homgeneous elements and $a+b$ lies in a graded ideal, then also $a,b$ lie in the ideal. Namely, if $a,b$ are of equal degree, then $b=-a$ provides counterexamples. However, it is true when the degrees of $a,b$ are different, because then $a$, $b$ are the homogeneous components of $a+b$ (resp. the other ones are $0$).
In order to understand this notion better, try to classify the graded ideals of $k[x]$, where $k$ is a field and $k[x]$ has the usual grading with $k[x]_n = k \cdot x^n$.
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