Let $F$ be a field and consider $F[x, y]$. Define $$(x, y) = \{\alpha x + \beta y \ | \alpha, \beta \in F[x, y]\}$$ Show that $(x, y)$ is an ideal of $F[x, y]$.
Here is my attempted proof:
Proof: Choose $a, b \in (x, y)$. Then $a = \alpha x + \beta y$ and $b = \xi x + \gamma y$ where $\alpha, \beta, \xi, \gamma \in F[x, y]$. We first show that $(x, y)$ is a subgroup of $F[x, y]$.
Observe that \begin{align*} a-b &= \alpha x + \beta y - (\xi x + \gamma y) \\ & = \alpha x + \beta y - \xi x - \gamma y \ \ \ \text{ because $F[x, y]$ is an abelian group } \\ &=(\alpha - \xi)x + (\beta - \gamma)y \ \ \text{ since $F[x, y]$ is an abelian group and a ring} \\ &\in (x, y) \end{align*}
So $(x, y)$ is a subgroup of $F[x, y]$ (by the subgroup test). Now choose $r \in F[x, y]$ and $a \in (x, y)$, then $a = \alpha x + \beta y$ and \begin{align*}r\cdot a &= r \cdot (\alpha x + \beta y) \\ &=r(\alpha x) + r(\beta y) \ \ \text{ by distributivity of $\cdot$ in $F[x, y]$} \\ &= (r\alpha)x +(r\beta)y \ \ \text{ by assosciativity of $\cdot$ in $F[x, y]$} \\ &\in (x, y) \end{align*}
Similarly one can show that $a \cdot r \in (x, y)$. Thus $(x, y)$ is an ideal of $F[x, y]$. $\square$
Firstly is my proof correct?
If so I don't see why we need $F$ to be a field for this to work, I haven't used any facts other than $F[x, y]$ being a ring to prove that $(x, y)$ is an ideal of $F[x, y]$. So if $R$ is a ring, then does it also hold that $(x, y)$ is an ideal in $R[x, y]$?
Your proof looks very close to correct! You also need to show that $(x,y)$ is nonempty. Indeed, it still works if $F$ is an arbitrary ring.