## 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.