Simultaneous congruences

Question

Let $\mathbb K$ be a finite field and $\mathbb K[x, y]$ the polynomial ring in the commuting indeterminates $x$ and $y$. Consider the factor ring $\mathbb K[x, y]/\langle x^3, y^3\rangle $. Can we find and element $\alpha\in\mathbb K[x, y]/\langle x^3, y^3\rangle $ such that: $$\begin{array} \alpha \alpha \equiv x \bmod \langle y\rangle \\ \alpha \equiv y \bmod \langle x+y\rangle \\ \alpha \equiv y \;\bmod \langle x\rangle \end{array}$$

Answer 1

$\alpha$ is a residue class of a polynomial $f$ from $k[x,y]$ modulo the ideal $(x^3,y^3)$. Then you can write $$f=f_1+f_2x+f_3y+f_4x^2+f_5xy+f_6y^2+f_7x^2y+f_8xy^2+f_9x^2y^2$$ with $f_i\in k$, or simpler $f=(f_1,\dots,f_9)$. (Note that $f$ is written in the canonical basis of the $k$-vectorspace $k[x,y]/(x^3,y^3)$ which I considered in some order.)

The first relation become $f-x-gy\in(x^3,y^3)$. Now write $g$ as before and find $$(f_1,f_2-1,\dots,f_9)=(0,0,g_1,0,g_2,g_3,g_4,g_5,g_7).$$ This shows that $f_1=f_4=0$ and $f_2=1$.

I leave you the pleasure to continue and find out that there is no such $\alpha$. (I hope my calculations are correct.)