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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?

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.

guestfromthepast
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    I'm confused about your last sentence, but there's no reason that an ideal has to be finitely generated in either the homogeneous or non-homogeneous case. For example take $k[x_1, x_2, \ldots]$ and take the ideal $(x_1, x_2, \ldots)$. – user148177 Oct 16 '14 at 08:08

2 Answers2

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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.

Ferra
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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.

Andrew Dudzik
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  • Does this proof work if you start with an uncountable set of generators? How do you get an ascending chain whose union is the ideal generated by all the generators in that case? – Callus - Reinstate Monica Mar 14 '17 at 18:41
  • @Callus Yes it does. Here is an elaboration: We inductively construct a chain of ideals contained in $I$. Let $I_0$ be the zero ideal, and, if $I_n$ does not equal $I$, then one of our generators $a_n$ must not be in $I$. Set $I_{n+1} = I + (a)$. Since this $I_{n+1}$ is strictly bigger than $I_n$, the notherian hypothesis tells us that this process must terminate, i.e. for some $n$, $I_n = I$. – Andrew Dudzik Mar 14 '17 at 23:02