subring not graded

Question

Definition 1. A pair $(A, \{A_{i}\}_{i\in\mathbb{Z}})$ is a graded ring if $A$ is a commutative ring with unit, and $\{A_{i}\}_{i\in\mathbb{Z}}$ is a family of $\mathbb{Z}$-submodules of $A$, such that

(1) $A = \bigoplus_{i}A_{i}$ as $\mathbb{Z}$-module.

(2) $\forall i$, $j$, $A_{i}A_{j}\subseteq A_{i + j}$.

Definition 2. Let $A = \bigoplus_{i}A_{i}$ be a graded ring. A graded ring $B = \bigoplus_{i}B_{i}$ is a graded subring of $A$, if $B$ is a subset of $A$, and the inclusion satisfies $\forall i$, $B_{i}\subseteq A_{i}$.

Remark. The definition of graded subring above is equivalent to that, a subring $B$ of $A$ is graded if and only if $B = \bigoplus_{i}(B\cap A_{i})$ as $\mathbb{Z}$-module. It seems that the only difference between subrings and graded subrings is whether $B\subseteq \bigoplus_{i}(B\cap A_{i})$ or not, i.e. closed or not in homogeneous decomposition.

Question. Is there an example of a subring of a graded ring that is not graded?

Answer 1

Take $A$ to be a polynomial ring $k[x]$ with the usual grading $x^i \in A_i$ and take $B$ to be the subring generated by, say, $x^2 + x$. $B$ does not contain any homogeneous polynomials of degree $1$ or $2$ and so is not equal to $\bigoplus_i (B \cap A_i)$.

Answer 2

Take $A = k[x,y]$ and $B = k[x]$, with $A_0 = k$, $y \in A_1$ and $x+y \in A_2$. In other words, we take the total degree grading on $k[z^2,y]$, and define the graded components of $A$ as the preimages of those of $k[z^2,y]$, under the $k$-isomorphism from $A$ to $k[z^2,y]$ defined by $x \mapsto z^2-y$ and $y \mapsto y$.

$x \in B$ can be written uniquely as a sum of homogeneous elements of different degrees, namely $x+y$ and $-y$, but these elements are not in $B$. So the grading of $A$ does not restrict to $B$. This is an example where $A = B[y]$ is the ring of polynomials over $B$. In particular, $B$ is integrally closed in $A$.