Answer to Question #160678 in Abstract Algebra for DHRUV PANDIL

Question

Let f is a field. Then prove that (0) is a prime ideal in fm

Answer 1

We will prove that $(0)$ is the maximal proper ideal and this will imply that it is prime. Suppose that $(0)$ is not maximal, then $\exists \mathfrak{m}, (0)\subsetneq \mathfrak{m}$ a maximal ideal. Then, $\exists a\in \mathfrak{m}, a\neq 0$, but this implies that $\forall b \in F, b\in \mathfrak{m},$ as $(b\cdot a^{-1})\cdot a \in \mathfrak{m}$ and thus $b\in \mathfrak{m}$ ($F$ is a field so for any non-zero element there is an inverse element). Therefore $\mathfrak{m} = F$ and so it is not proper. So we conclude that $(0)$ is a maximal proper ideal and thus is prime.