Lemma 53.15.1. Let $k$ be an algebraically closed field. Let $k \subset A$ be a ring extension such that $A$ has exactly two $k$-sub algebras, then either $A = k \times k$ or $A = k[\epsilon ]$.
Proof. The assumption means $k \not= A$ and any subring $k \subset C \subset A$ is equal to either $k$ or $A$. Let $t \in A$, $t \not\in k$. Then $A$ is generated by $t$ over $k$. Hence $A = k[x]/I$ for some ideal $I$. If $I = (0)$, then we have the subalgebra $k[x^2]$ which is not allowed. Otherwise $I$ is generated by a monic polynomial $P$. Write $P = \prod _{i = 1}^ d (t - a_ i)$. If $d > 2$, then the subalgebra generated by $(t - a_1)(t - a_2)$ gives a contradiction. Thus $d = 2$. If $a_1 \not= a_2$, then $A = k \times k$, if $a_1 = a_2$, then $A = k[\epsilon ]$. $\square$
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