Definition of homogeneous ideal

Question

I'm a little confused about the definition of a homogeneous ideal. I have the following two definitions:

• An ideal $I\subset k[X_{0}, \dots, X_{n}]$ is homogeneous if $I$ is generated by (finitely many) homogeneous polynomials.

• An ideal $I\subset k[X_{0}, \dots, X_{n}]$ is homogeneous if $I$ can be generated by homogeneous polynomials.

So does a homogeneous ideal have to be an ideal generated only by finitely many homogeneous polynomials or do we allow the infinite case?

I know that every ideal can be generated by finitely many when $k$ is a field, but an ideal, which we have generated by an infinite number of homogeneous polynomial, is not necessarily generated by a finite number of homogeneous polynomials.. right?

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To clarify my last sentence: I had an exercise where $I\subset k[x_{1}, \dots, x_{n} ]$ is an ideal and $I^{h}$ was the ideal generated by $\lbrace f^{h};f\in I \rbrace$ where $f^{h}$ is the homogenization of $f$ i.e. it is a homogeneous polynomial.

The exercise was to show that $I^{h}$ is a homogeneous ideal. From the second definition above, this is "obvious" a homogeneous ideal (since it is generated by homogeneous polynomials). But from the first definition we actually have to show that it can be generated by a finite number of homogeneous polynomials. The solution I was given, was not a trivial solution.

Answer 1

In a notherian ring, an ideal with an infinite set of generators is generated by some finite subset of those generators (just keep taking more non-redundant ones until Emmy tells you to stop). So this distinction is not important. If we have generators that are homogeneous, then we have a finite set of generators that are homogeneous.

Maybe I'm missing something, but if somebody solved the problem you mention in a non-trivial way, it's possible that they were inadvertently re-proving something like the Hilbert basis theorem.

Answer 2

Let's prove the following statement: an ideal $I\subseteq k[X_0,\ldots,X_n]$ has a set of homogeneous generators iff whenever $f=\sum_if_i\in I$ where $f_i$ are the homogeneous components of $f$, then $f_i\in I$ for all $i$.

In fact, suppose $I$ has a set $\{f_{\lambda}\}_{\lambda\in \Lambda}$ of homogeneous generators. Now take $g\in I$ and let $g=\sum_i g_i$ be its decomposition in homogeneous components. Then $g=\sum_{j=1}^m h_jf_{\lambda_j}$ for some $m\in \mathbb N$ and $h_j\in k[X_0,\ldots,X_n]$. Now decompose $\sum h_jf_{\lambda_j}$ in its homogeneous components. If you think a bit about it, you will see that since the $f_{\lambda_j}$ are homogeneous, every homogeneous component of $\sum h_jf_{\lambda_j}$ will be of the form $\sum_{k\in K}h'_k f_{\lambda_k}$ where $K\subseteq \{1,\ldots,m\}$. Since this decomposition is unique, this means that every $g_i$ has to coincide with an expression of the type $\sum_{k\in K}h'_k f_{\lambda_k}$, which implies that $g_i\in I$.

Conversely, suppose that $f=\sum_if_i\in I$ implies $f_i\in I$ forall $i$. Take a set $\{g_1,\ldots,g_m\}$ of generators for $I$. Then also the set $\{h\in k[X_0,\ldots,X_n]\colon h \mbox{ is a homogeneous component of some }g_i\}$ is obviously a set of generators for $I$. Note that this is a finite set!

In particular we showed that if $I$ has a set of homogeneous generators, than it has a FINITE set of homogeneous generators. So the answer to you question is: the two definitions are equivalent.