I am reading Balmer's paper "Spectra, Spectra, Spectra" regarding the spectrum of tensor-triangulated categories. I think I am missing something obvious when he states that all ideals are radical as soon as the tensor triangulated category is rigid.
Recall that a tensor-triangulated category $\mathcal T$ is rigid when there is an exact functor $D\colon \mathcal T^{op}\to \mathcal T$ such that for each $x,y,z\in \mathcal T$ $$ \text{Hom}_{\mathcal T}(x\otimes y, z)\cong \text{Hom}_{\mathcal T}(x,D(y)\otimes z) $$ as abelian groups.
To show: If $\mathcal T$ is rigid, then every ideal $\mathfrak a$, i.e., a triangulated subcategory which is closed under tensor products by $\mathcal T$ and closed under summands, is radical, i.e., for each $x\in \mathcal T$ such that $x^{\otimes n}\in \mathfrak a$ for some $n>0$, then $x\in\mathfrak a$.
The following result may prove useful: In a tensor-triangulated category $\mathcal T$, every ideal is radical iff for each $x\in \mathcal T$, $x\in \langle x\otimes x\rangle$ (the ideal generated by $x\otimes x$).
What I was thinking is to look at Hom$_{\mathfrak a}(x\otimes x,x\otimes x)\cong$Hom$_{\mathcal T}(x,D(x)\otimes(x\otimes x))$ or playing around with some other of these isos, but nothing is jumping out at me to suggest that $\mathfrak a$ is radical. Any hints would be appreciated.
The proof was in Balmer's older 'Supports and filtration...' (Prop. 2.4). It is quite simple: by the unit-counit relation, every rigid x is a direct summand of $x \otimes Dx \otimes x$. So if $x \otimes x$ belongs to an ideal then so does $x$. Etc.