I am currently trying to show that the ideal $I = \langle x, y-1 \rangle$ is a prime ideal in $R = k[x,y]/\langle xy \rangle$ (for some field $k$).
My first thought was to rewrite the ideal as follows:
$\langle x, y-1 \rangle = \{ \alpha x + \beta(y-1) + \gamma (xy-x) : \alpha, \beta, \gamma \in R\}$
Since $xy=0$ in the ring, we rewrite this as:
$I = \{ (\alpha - \gamma) x + \beta y - \beta : \alpha, \beta, \gamma \in R\}$
But I quickly get stuck here. I know I want to show that if $fg \in I$, then $f \in I$ or $g \in I$, but how would I do this in this case? I can't see any nice way of doing it.
If this the correct way to go about this (and if so, could I have a nudge in the right direction), or have I gone the wrong way about doing it?
Thanks in advance!
Joseph Curwen is right. The trick lies in the correspondence theorem.
Lemma.- Let $f: R \to S$ be a surjective ring homomorphism, $p$ a prime ideal in $R$ containing the kernel of f. Then $f(p)$ is a prime ideal in $S$.
Proof is easy.
Now apply this lemma to the factor homomorphism $k[x,y] \to k[x,y]/<xy>$. You only need to show that the ideal $<x,y-1>$ of $k[x,y]$ is prime. For this, recall that any polynomial in two variables can be written uniquely as:
$$f(x,y) = \sum a_{ij} x^i (y-1)^j$$
for some $a_{ij} \in k$.
EDIT:
This proof is incorrect. My apologies. The kernel is not contained in the prime ideal. Here is an idea.
We denote by $I$ the ideal generated by $x$ and $y-1$ in the quotient ring $k[x,y]/<xy>$. Denote by $\pi: k[x,y] \to k[x,y]/<xy>$ the quotient map. By the third isomorphism theorem for rings, we have:
$$\frac{k[x,y]/<xy>}{I} = \frac{k[x,y]/<xy>}{\pi^{-1}(I)/<xy>} \cong k[x,y]/\pi^{-1}(I)$$
So one only needs to show that $k[x,y]/\pi^{-1}(I)$ is an integral domain.