How I prove that $\frac{K[x, y]}{ \langle x^2 - x, y^2 - y\rangle} \cong K \oplus K\oplus K\oplus K$? How I prove that $\frac{K[x, y]}{ \langle x^2 - x, y^2 - y\rangle} \cong K \oplus K\oplus K\oplus K$?
 A: We shall be using this result .
Observe that 
$$ \frac{\frac{K[x, y]}{\langle x^2 - x\rangle}}{ \langle  {\overline y}^2 - {\overline y}\rangle}\cong \frac{K[x, y]}{ \langle x^2 - x, y^2 - y\rangle} $$
We know that if $I$ is an ideal of a ring $R$ then $\frac{R[y]}{IR[y]}\cong \frac{R}{I}[y]$. Take $R=K[x]$ and $I=\langle x^2 - x\rangle$ $\implies \frac{K[x, y]}{\langle x^2 - x\rangle}\cong \frac{K[x]}{\langle x^2 - x\rangle}[y].$
So,
$$\frac{\frac{K[x]}{\langle x^2 - x\rangle}[y]}{\langle  {y}^2 - {y}\rangle}\cong\frac{\frac{K[x, y]}{\langle x^2 - x\rangle}}{ \langle  {\overline y}^2 - {\overline y}\rangle}\cong \frac{K[x, y]}{ \langle x^2 - x, y^2 - y\rangle}$$
Now use the fact for any ring (this fact is true by Chinese Reminder Theorem) $$\frac{R[x]}{\langle x^2-x\rangle}\cong R\oplus R$$
twice. 
Once you take take $R=\frac{K[x]}{\langle x^2 - x\rangle}$ and have
$$ \frac{K[x, y]}{ \langle x^2 - x, y^2 - y\rangle}\cong \frac{\frac{K[x]}{\langle x^2 - x\rangle}[y]}{\langle  {y}^2 - {y}\rangle}\cong \frac{K[x]}{\langle x^2 - x\rangle}\oplus \frac{K[x]}{\langle x^2 - x\rangle}$$
 Then for the second time you take $R=K$ and have
$$\frac{K[x]}{\langle x^2 - x\rangle}\cong K\oplus K$$
