Let $O_K$ be the ring of integers of an algebraic number field $K$ and let $\mathfrak{p}$ be a prime in $O_K$. For a field extension $L/K$, we consider the ring of integer $O_L$ in $L$ and the ideal $\mathfrak{p} O_L$ of $O_L$. If $[L:K]< \infty$, then we have $$\mathfrak{p} O_L=\mathfrak{p}_1^{e_1} \mathfrak{p}_2^{e_2} \cdots \mathfrak{p}_n^{e_n},$$ where $\mathfrak{p}_i$ are distinct prime ideals of $O_L$.
Then $\mathfrak{p}$ ramifies in $L$ if some of $e_i>1$.
My question:
Why the above condition ($e_i>1$ for some $i$) is equivalent to say that $O_L/\mathfrak{p} O_L$ has a non-zero ${\color{blue}{nilpotent}}$ element ?
Answer:
Let $e_1=2>1$ and $\mathfrak{p} O_L=\mathfrak{p}_1^2$, where $\mathfrak{p}_1$ is the prime ideal in $O_L$.
How does this imply that $O_L/\mathfrak{p} O_L$ has non-zero nilpotent element ?
Best Answer
By the Chinese Remainder Theorem we have that
$$ \frac{O_L}{\mathfrak{p}_1^{e_1} \mathfrak{p}_2^{e_2} \cdots \mathfrak{p}_n^{e_n}} \cong \frac{O_L}{\mathfrak{p}_1^{e_1}} \times \cdots \times \frac{O_L}{\mathfrak{p}_n^{e_n}}.$$
If any $e_i$ is greater than $1$, then the quotient ring $\frac{O_L}{\mathfrak{p}_i^{e_i} }$ has a nonzero nilpotent element, namely, any element of $\mathfrak{p_i} - \mathfrak{p}_i^{e_i}$ modulo $\mathfrak{p}_i^{e_i}$. So the product also has a nonzero nilpotent element.
If all the $e_i$ are equal to $1$, then $\frac{O_L}{\mathfrak{p}_i^{e_i}}= \frac{O_L}{\mathfrak{p}_i}$ is a field. And a product of fields does not have any nonzero nilpotent elements.