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Let $R$ be a commutative ring. The Prime Avoidance lemma says that if an ideal $I\subseteq R$ is not contained within the prime ideals $p_1,p_2$, then it is not contained within $p_1\cup p_2$ either.

Let $R$ be the ring $\Bbb{C}[x,y]$, and let $I=(x,y), p_1=(x+y)$ and $p_2=(x-y)$. Then clearly, $(x,y)\not\subseteq (x+y),(x-y)$. However, it is included in the union of $(x+y)$ and $(x-y)$, which is precisely $(x,y)$. Isn't this a contradiction? Where am I going wrong?

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    $\begingroup$ I don't think $(x+y) \cup (x-y) = (x,y)$... It is true that $(x+y) + (x-y) = (x,y)$, i.e., the ideal generated by $(x+y) \cup (x-y)$ is $(x,y)$. $\endgroup$ – Viktor Vaughn Jul 4 '17 at 1:21
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    $\begingroup$ Ah okay. Thanks. I was confusing $p_1\cup p_2$ with $p_1+p_2$. Thanks $\endgroup$ – user67803 Jul 4 '17 at 1:57
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As I said in the comments, your mistake is that $(x+y) \cup (x-y) \neq (x,y)$; rather this is true for the ideal generated by $(x+y) \cup (x-y)$, namely, $(x+y) + (x-y) = (x,y)$.

One can show $(x+y) \cup (x-y) \neq (x,y)$ as follows. For contradiction, assume $(x+y) \cup (x-y) = (x,y)$. Then WLOG $x \in (x+y)$, so $y = x+y - x \in (x+y)$, hence $(x+y) = (x,y)$. A principal ideal has height $1$ by Krull's Hauptidealsatz, but $(0) \subsetneq (x) \subsetneq (x,y)$ is a chain of primes of length $2$, so $(x,y)$ has height $\geq 2$, contradiction. (One could also argue directly by writing $x = (x+y)f(x,y)$ and deriving a contradiction.)

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