Graded rings: what does $\mathbb{Z}[y]/(y, y^{2n + 1})$ with $y$ of degree $2$ mean?

Question

I do not understand what $\mathbb{Z}[y]/(2y, y^{2n + 1})$ with $y$ of degree $2$ means.

If I read the Wikipedia page right, the graded ring $\mathbb{Z}[y]$ is the set of all polynomials in $y$ with coefficients in $\mathbb{Z}$ with the natural multiplication, $(2y, y^{2n + 1})$ is the ideal generated by $(2y, y^{2n + 1})$.

Then $\mathbb{Z}[y]/(y, y^{2n + 1})$ is $\bigoplus(R_i + I)/I$ with the canonical multiplication, where $R$ is all polynomials of the form $z y^i$, $I$ is the ideal $(y, y^{2n + 1})$ and $R_i + I = \{a + b \ | \ a \in R_i, b \in I\}$.

Question 2: Wouldn't that mean that $(R_i + I)/I \cong R$ because $[a + b] = [a]$ in $(R_i + I)/I$?

Question 3: What does $y$ of degree 2 mean?

Edit: A bit more context may help:

I need to find an isomorphism of graded rings

$$H^*(\mathbb{R}P^{2n}; \mathbb{Z}) \cong \mathbb{Z}[y]/(2y, y^{2n + 1})$$

Answer 1

Of course, $y^{2n+1}$ lies in the principal ideal generated by $y$, and hence is a redundant (or superfluous) generator. Also, the corresponding quotient ring is clearly isomorphic to $\mathbb{Z}$.