Lemma 15.9.9. Let $I$ be an ideal of a ring $A$. Let $A \to B$ be an integral ring map. Let $b \in B$ map to an idempotent in $B/IB$. Then there exists a monic $f \in A[x]$ with $f(b) = 0$ and $f \bmod I = x^ d(x - 1)^ d$ for some $d \geq 1$.
Proof. Observe that $z = b^2 - b$ is an element of $IB$. By Algebra, Lemma 10.38.4 there exist a monic polynomial $g(x) = x^ d + \sum a_ j x^ j$ of degree $d$ with $a_ j \in I$ such that $g(z) = 0$ in $B$. Hence $f(x) = g(x^2 - x) \in A[x]$ is a monic polynomial such that $f(x) \equiv x^ d(x - 1)^ d \bmod I$ and such that $f(b) = 0$ in $B$. $\square$
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Your email address will not be published.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)