Clearly if a Jordan algebra has nilpotent element, the algebra is non-simple.
I suppose the converse is not true and there are non-simple Jordan algebras without nilpotent elements, but there might be a theorem that I'm missing. In particular I'm working with infinite dimensional algebras (so a nilpotent element do not guarantee the existence of a maximal idempotent element, if that's needed).
Could you please tell me if that is the case? Sorry if this question is trivial, but really is not my expertise.
Fix a set of indexes $I$ (possibly uncountable) and consider $B=\{a_i\}_{i\in I}$. Consider the $\mathbb{Z}_2$ algebra $A$ with basis $B$ and non-zero products given by $a_i^2=a_i$. We can write $x\in A$ as a linear combination of say $a_{i_1},...,a_{i_k}$ and $y\in A$ as a combination of $a_{j_1},...,a_{j_l}$, either way we consider both $x$ and $y$ as a combination of $a_{i_1},...,a_{i_k},a_{j_1},...,a_{j_l}$ by adding a zero as a coefficient in case some element does not appear in the expression and label them as $a_1,...,a_p$ (this will simplify the notation).
From that we have $(\sum_{r=1}^p x_{r}a_{r})(\sum_{s=1}^p y_{s}a_{s})=\sum\sum x_{r}y_{s}a_{r}a_{s}=\sum x_{r}y_{r}a_{r}^2$. In particular, if $x=y$ it follows that $x^2=x$. Therefore the only nilpotent element is $0$.
The vectos space $\langle a_1\rangle$ is also an ideal for $(\sum_{r=1}^p x_{r}a_{r})a_1=\delta_{1,x}a_1$ where $\delta_{1,x}=0$ if $a_1$ does not appear in the expression of $x$ and $\delta_{1,x}=x_1$, the coefficient of $a_1$ in the expression of $x$. Either way it belongs to the space generated by $a_1$. Since $A$ is commutative, it is an ideal. The Jordan identity is immediate.