Show $\mathbb {R}[x,y]/(y^2-x, y-x)$ is not an integral domain 
Let $\mathbb{R}[x,y]$ denote the polynomial ring in two variables $x$, $y$ over $\mathbb{R}$, and let $I = (y^2-x,y-x)$ be the ideal generated by $y^2-x$ and $y=x$. Show that $$\mathbb{R}[x,y]/I$$ is not an integral domain. 

To be honest, I have no idea to solve this question. I am thinking that if the $\mathbb{R}[x,y]/I$ is not an integral domain, $I$ is neither a prime ideal not a maximal ideal. But I don't know how to prove that. Could you please help me? Thank you very much!
 A: $\mathbb R [x,y] / (y-x) \cong \mathbb R[y]$ (trivially), and factoring by $(y^2-x)$ in the former means factoring by $(y^2-y)$ in the latter. And $\mathbb R[y]/(y^2-y)$ is not a domain, can you see why?
A: 
The golden rule. In a factor ring $R/I$ we have $a=0\bmod I$ iff $a\in I$ (where $a\in R$ ). 

In your example $y^2=x\bmod I$ and $y=x\bmod I$. This gives us $y^2=y\bmod I$, so $y(y-1)=0\bmod I$, and this suggests that $y\bmod I$ and $y-1\bmod I$ are zero divisors in $R/I$. (Here $R=\mathbb R[x,y]$.)
However we have to check that they are not zero. For instance, suppose $y=0\bmod I$. Then, from the golden rule, $y\in I$, that is, $y\in (y^2-x,y-x)$. Now write $$y=(y^2-x)f(x,y)+(y-x)g(x,y),$$ and for $x=y=1$ we get $1=0$, a contradiction. (Do the same for $y-1\bmod I$.)
A: Hint: think of your ideal as $\langle y^2-x\rangle+\langle y-x\rangle$. So let us quotient $\mathbb R[x,y]$ by $\langle y-x\rangle$ first. What do we get? In this new ring, $y=x$, so plug that in to the other ideal to reduce our computation to a ring with only one variable! It should be much easier now.
Work through the above stuff and let me know if you want more hints
A: The ideal $I=(y^2-x,x-y)$ is not prime, since $y^2-y=(y^2-x)+(x-y)$ lies in $I$, but neither $y$ nor $y-1$ lies in $I$. Indeed, writing
$$y=p(x,y)(x^2-y)+q(x,y)(x-y)$$ and setting $x=y=1$ we get $1=0$, which is impossible. The same argument holds for $y-1$. Geometrically, $k[x,y]/(x^2-y,y-x)$ should codify the intersection of the parabola $x^2=y$ with the line $y=x$. This is just the points $(1,1)$ and $(0,0)$, and this corresponds to an isomorphism of such quotient with the non domain $k\times k$. 
One should not be carried away by simplifications, however. If we had alternatively taken the tangent line $y=0$ at $(0,0)$, the resulting quotient $k[x,y]/(x^2-y,y)=k[x]/(x^2)=k[\varepsilon]$ is still not a domain although the intersection is "one point" (it isn't: it is a double point, if you may): this is because the tangent intersects the parabola with multiplicity two. Note that the resulting ring is no longer disconnected (that is, a direct product $k\times k$), and indeed this codifies such tangential intersection: an element of the ring $a+b\varepsilon$ is specified by a scalar $a$ and a small perturbation $b\varepsilon$ with $\varepsilon^2=0$. 
