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What is the intuition (hopefully, geometric) behind these basic facts about homogeneous ideals? An ideal $I$ in $S$ is homogeneous if an element $f = \sum_{n \ge 0} f_n$ of $S$ lies in $I$ if and only if each homogeneous part $f_n$ lies in $I$. Here, we let $S = \bigoplus_{n \ge 0} S_n$ be an $\mathbf{N}$-graded ring.

  • An ideal generated by homogeneous elements is a homogeneous ideal.

  • A homogeneous ideal $I$ in $S$ is prime if and only if it is a proper ideal and $$fg \in I \implies f \in I \text{ or }g \in I$$for homogeneous $f, g \in S$.

  • The kernel of $\mathbf{N}$-graded rings is a homogeneous ideal.

  • For any homogeneous ideal $I$ of $S$ there is a natural $\mathbf{N}$-grading on $S/I$.

I am not an algebraic geometrer by trade, but I need to use these results.

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3 Answers 3

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If you have $S=K[x_1,\ldots, x_n]$ with the usual grading, then ideals will correspond to closed subsets of $\mathbb{A}^n$. An ideal $I$ is homogeneous if and only the zero set $V(I)$ satisfies the condition that if $(a_1,\ldots a_n)\in V(I)$ and $\lambda\in K\backslash 0$ then $(\lambda a_1\ldots, \lambda a_n)\in V(I)$. If you think of the natural projection $\mathbb{A}^n\backslash 0\rightarrow \mathbb{P}^{n-1}$, then these are exactly the inverse images of the Zariski closed sets in $\mathbb{P}^{n-1}$.

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One thing to grok is that, in my opinion, graded rings are a hack. Occasionally graded rings as defined are of interest, but usually, at least in the fields I'm familiar with, the inhomogeneous elements don't play a role at all, and their presence is only a trick so that you can leverage the existing body of knowledge of commutative algebra.

The actual algebraic structure you care about is the essentially algebraic one made up out of the graded components, and the arithmetic with them (i.e. addition should only be defined for things of the same grading). This can be phrased in terms of preadditive categories, with one object per grading. A ring, for example, is a preadditive category with one object.

(also, I believe this foundation for graded rings wasn't known for a long time, so the version you cite is less of a trick and more of the only thing people had to work with)

The intuition behind the theorems you list is simply that when using this hack, the usual definitions of things have some relevance to what you want for graded rings (e.g. that a homogeneous prime ideal when represented as an ordinary ideal is actually prime). That is, the inclusion of the inhomogeneous things don't get in the way of the theory working in a simple way.

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  • $\begingroup$ Note that this viewpoint is now formalized in the theorem prover Lean, which provides a plethora of helper theorems to move between the "extrinsic approach" given by a sequence of abelian groups with multiplication, and the "intrinsic approach" which is a ring separated into the direct sum of homogeneous components. $\endgroup$
    – Trebor
    Oct 9, 2023 at 14:41
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Let $X$ be an affine variety and $A$ its coordinate ring. Then a $ \mathbb{Z} $ grading on $A$ is the same thing as a $ \mathbb{G}_m = k^{\times} $ action on $X$. If $Y$ is a closed subvariety cut out by the ideal $I$, then $I$ is graded if and only if $Y$ is invariant under the torus action. This should give you some intuition for graded ideals: they are exactly those which cut out torus invariant subvarieties.

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    $\begingroup$ There's no need to pass to geometry yet. The $\mathbb{Z}$ grading on $S$ is the same as the decomposition of $S$ into its isotypic components as a representation of $\mathbb{G}_m$ acting by multiplying all the $x_i$ simultaneously, and all the facts above are really facts about ideals that are subrepresentations (of an abelian group, all of whose irreps are 1-dimensional). Of course geometry will eventually yield insight; if it didn't there would be no GIT theory! $\endgroup$ Aug 4, 2015 at 3:29
  • $\begingroup$ @AlexanderWoo: well said! $\endgroup$ Aug 4, 2015 at 12:28

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