An ideal of R[x] is homogeneous if and only if it has a generating set consisting only of homogeneous polynomials

Let R be a ring. An ideal I \subseteq R[x_1,\ldots,x_n] is called homogeneous if whenever \alpha \in I, each homogeneous component of \alpha is also in I. Prove that an ideal is homogeneous if and only if it has a generating set consisting only of homogeneous polynomials.


Let I be a homogeneous ideal. There exists a generating set A of I; let \overline{A} denote the set of all homogeneous components of elements in A. Certainly we have A \subseteq (\overline{A}), so that I \subseteq (\overline{A}). Since I is homogeneous, we also have \overline{A} \subseteq I, so that I = (\overline{A}). Thus I has a generating set consisting only of homogeneous polynomials.

Conversely, suppose I = (A) is generated by homogeneous polynomials; let A = \{a_t\}_T. We will show that for every \alpha \in I, each homogeneous component of \alpha is in I by induction on the degree of \alpha. For the base case, let \alpha \in I have minimal degree k. We have \alpha = \sum_t r_t a_t, where each a_t is homogeneous and in I. since \alpha has minimal degree, in fact the degree of each a_t in this decomposition of \alpha has degree k, so that \alpha is itself homogeneous of degree k. In particular, every homogeneous component of \alpha is in I.

For the inductive step, suppose that for some n \geq k, if \alpha \in I is a polynomial of degree at most n, then every homogeneous component of \alpha is in I. Now let \alpha \in I have degree n+1. Write \alpha = \sum_t r_ta_t + \sum_t s_ta_t, where \sum_t s_ta_t consists only of the degree n+1 polynomials in A. Note that \sum_t r_ta_t is in I, and consists precisely of those homogeneous components of \alpha of degree at most n. Thus each homogeneous component of \alpha of degree at most n is in i. Moreover, since \sum_t s_t a_t = \alpha - \sum_t r_ta_t, the homogeneous component of degree n+1 is in I. So every homogeneous component of \alpha is in I.

By induction, I is a homogeneous ideal.

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