Show that there is no proper ideal of finite presentation of $R$. Let  $(A, M) $ be a local ring and $E$ a $(A/M)$-vector space infinite dimension. The trivial ring extension of $ A$ by $E$ (also called the idealization of $E$ over $A$) is the ring $R := A ∝ E $ whose underlying group is $A×E $ with with addition given by : $$ (a,e)+(a',e')=(a+a',e+e')$$ and multiplication given by : $$(a,e)(a′,e′) = (aa′,ae′ +a′e)$$
As ring $A$ is local so $R$ is local and its maximum ideal is $M∝E$.
{Problem} : 
Show that there is no proper ideal of finite presentation of $R$.
{ my effort}:
Let $I $ be a proper ideal of $R $ of finite presentation, then $I$ is finitely generated that implies :
$$
I=\sum_{i=1}^{n}{R (a_i,e_i)}=\sum_{i=1}^{n}{ (A ∝ E)(a_i,e_i)}=\sum_{i=1}^{n}{ (A a_i)∝ (Ae_i+a_i E)}=(\sum_{i=1}^{n}{ A a_i})∝ (\sum_{i=1}^{n}{ Ae_i+a_i E})=
(\sum_{i=1}^{n}{ A a_i})∝ (\sum_{i=1}^{n}{ Ae_i})
$$
Because  : $I\subset M∝E$ then $\sum_{i=1}^{n}{ A a_i}\subset M$ that implies $a_i\in M $ thus $a_i E=0$ because $E$ est un $(A/M)$ - vector space.
But I can not find a contradiction.
 A: Suppose an ideal $I$ of $R$ is generated by $(a_{1},e_{1}) , \dotsc , (a_{n},e_{n})$ as you say. We may assume that there are no "redundancies" among the $(a_{i},e_{i})$ as follows: if we can write some $(a_{i},e_{i})$ as an $R$-linear combination of $(a_{i'},e_{i'})$ for $i' \ne i$, then we remove $(a_{i},e_{i})$.
We show that the "module of $R$-linear relations between the $(a_{i},e_{i})$" is not finitely generated. This module of relations is the kernel of the map $\varphi : R^{\oplus n} \to R$ sending the $i$th basis element to $(a_{i},e_{i})$. Observe that a vector $$ (\mathbf{b},\mathbf{f}) \in A^{\oplus n} \propto E^{\oplus n} =R^{\oplus n} $$ is contained in $\ker \varphi$ if and only if $\mathbf{b} \in \ker(A^{\oplus n} \to R)$ where this map is the restriction of $\varphi$. In particular, $0 \propto E^{\oplus n} \subseteq \ker \varphi$. Note also that if $(\mathbf{b},\mathbf{f}) \in \ker \varphi$, then $\mathbf{b} \in M^{\oplus n}$ for all $i$: if the $i$th coordinate of $\mathbf{b}$ is contained in $A \setminus M$, then the $i$th coordinate of $(\mathbf{b},\mathbf{f})$ is a unit, which would allow us to solve for $(a_{i},e_{i})$ in terms of the others; this is not possible by our assumption that there are no "redundancies").
So suppose for the sake of contradiction that $$ (\mathbf{b}_{1},\mathbf{f}_{1}) , \dotsc , (\mathbf{b}_{m},\mathbf{f}_{m}) \in \ker \varphi $$ are vectors that generate $\ker \varphi$ (here $\mathbf{b}_{j} \in A^{\oplus n}$ and $\mathbf{f}_{j} \in E^{\oplus n}$). Consider an $R$-linear relation $$ (c_{1},g_{1}) \cdot (\mathbf{b}_{1},\mathbf{f}_{1}) + \dotsb + (c_{m},g_{m}) \cdot (\mathbf{b}_{m},\mathbf{f}_{m}) $$ with $(c_{j},g_{j}) \in R$. The second coordinate of the above linear combination is $$ (c_{1}\mathbf{f}_{1} + \dotsb + c_{m}\mathbf{f}_{m}) + (\mathbf{b}_{1}g_{1} + \dotsb + \mathbf{b}_{m}g_{m}) $$ in $E^{\oplus n}$; the first expression is contained in $\operatorname{Span}(\mathbf{f}_{1},\dotsc,\mathbf{f}_{m})$ and the second expression is $0$ since $\mathbf{b}_{j} \in M^{\oplus n}$ for all $j$. Since $E$ is infinite-dimensional, we have that $\operatorname{Span}(\mathbf{f}_{1},\dotsc,\mathbf{f}_{m})$ does not generate $E^{\oplus n}$, contradiction.
Finally, any other presentation is not a finite presentation by Schanuel's lemma.
