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An ideal $I$ in $R\left[x_{1}, \ldots, x_{n}\right]$ is called a homogeneous ideal if whenever $p \in I$ then each homogeneous component of $p$ is also in $I$ . Prove that an ideal is a homogenecus ideal if and only if it may be generated by homogeneous polynomials, [Use inducticn on degrees to show the "if" implication. The following exercise shows that some care must be taken when working with polynomials over noncommutative rings $R$ (the ring operations in $R[x]$ are defined in the same way as for commutative rings $R,$ in perticular when considering polynomials as functions.

Solution

Verified$\hspace*{5mm}$ Let $\Bbb{R}$ be a ring and $I$ ideal in $\Bbb{R}$.

We suppose that $I$ is a homogeneous ideal. $A$ is generating set of $I$. $\overline{A}$ denote the set of all homogeneous component of elements in $A$.

$\hspace*{5mm}$From this we have: $A \subseteq ( \overline{A})$, and $I \subseteq ( \overline{A}) .$

But $I$ is homogeneous ideal and from it we have $I \supseteq ( \overline{A}) .$

From $I \subseteq ( \overline{A})$ and $I \supseteq ( \overline{A})$ we conclude $\mathbf{A = ( \overline{A})}$, and

I may be generated by a homogeneous polynomial.

$\hspace*{5mm}$ Suppose $A=\left \{ a_i \right \}$.

This part we prove by induction on $a$

For $n \geq k$: $\hspace*{3mm}$ If $a$ is a polynomial of degree at most n, then every component of a is in $I$.

For $n +1$:

$\hspace*{3mm}$ $a=\sum_{i}r_ia+\sum_{i} s_ia_i$ where $\sum_{i} s_ia_i$ consist only of degree $n+1$ polynomial in $A$.

$\hspace*{3mm}\sum_{i}r_ia$ consist of homogeneous components of $a$ at most degree $n$ and $\sum_{i}r_ia$ is in $I$.

$\hspace*{3mm}$Each homogeneous components of $a$ at most degree $n$ is in $I$

$\hspace*{7mm}\sum_{i} s_ia_i=a+\sum_{i}r_ia$ degree $n+1$ is in $I$.

$\hspace*{7mm}$ Every homogeneous components of $a$ is in $I$.

An Ideal is homogeneous ideal if and only if it may be generated by homogeneous polynomial.

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