Lemma 10.15.3. Let $R$ be a ring. Let $x \in R$, $I \subset R$ an ideal, and $\mathfrak p_ i$, $i = 1, \ldots , r$ be prime ideals. Suppose that $x + I \not\subset \mathfrak p_ i$ for $i = 1, \ldots , r$. Then there exists a $y \in I$ such that $x + y \not\in \mathfrak p_ i$ for all $i$.
Proof. We may assume there are no inclusions among the $\mathfrak p_ i$. After reordering we may assume $x \not\in \mathfrak p_ i$ for $i < s$ and $x \in \mathfrak p_ i$ for $i \geq s$. If $s = r + 1$ then we are done. If not, then we can find $y \in I$ with $y \not\in \mathfrak p_ s$. Choose $f \in \bigcap _{i < s} \mathfrak p_ i$ with $f \not\in \mathfrak p_ s$. Then $x + fy$ is not contained in $\mathfrak p_1, \ldots , \mathfrak p_ s$. Thus we win by induction on $s$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: