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

Let be a ring. An ideal is called *homogeneous* if whenever , each homogeneous component of is also in . Prove that an ideal is homogeneous if and only if it has a generating set consisting only of homogeneous polynomials.

Let be a homogeneous ideal. There exists a generating set of ; let denote the set of all homogeneous components of elements in . Certainly we have , so that . Since is homogeneous, we also have , so that . Thus has a generating set consisting only of homogeneous polynomials.

Conversely, suppose is generated by homogeneous polynomials; let . We will show that for every , each homogeneous component of is in by induction on the degree of . For the base case, let have minimal degree . We have , where each is homogeneous and in . since has minimal degree, in fact the degree of each in this decomposition of has degree , so that is itself homogeneous of degree . In particular, every homogeneous component of is in .

For the inductive step, suppose that for some , if is a polynomial of degree at most , then every homogeneous component of is in . Now let have degree . Write , where consists only of the degree polynomials in . Note that is in , and consists precisely of those homogeneous components of of degree at most . Thus each homogeneous component of of degree at most is in . Moreover, since , the homogeneous component of degree is in . So every homogeneous component of is in .

By induction, is a homogeneous ideal.

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