I am not sure if this is a trivial result. Is it true that if an abelian monoidal category $\tilde{\mathfrak{C}}$ admits a fibre functor then each object of the underlying abelian category $\mathfrak{C}$ is artinian and noetherian? Why? Or maybe a weaker result, are tannakian categories also artinian and noetherian? Why?
Thanks.
Let $F$ be the fibre functor. If $f:X\to Y$ is a monomorphism, but not an epimorphism, in $\mathfrak{C}$, then $Ff:FX\to FY$ is a monomorphism of vector spaces by exactness, but not an epimorphism since otherwise the identity map on $\text{cok}f$ would be sent to zero, contradicting faithfulness.
So if an object $X$ of $\mathfrak{C}$ has a chain $$0=X_0<X_1<\dots<X_n=X$$ of subobjects, each a proper subobject of the next, then applying $F$ we get a chain of monomorphisms, but not epimorphisms, of vector spaces $$0=FX_0\to FX_1\to\dots\to FX_n=FX,$$ which is impossible if $\dim FX<n$.
So if $\dim FX<\infty$ then $X$ cannot have an infinite chain, descending or ascending, of subobjects, so $X$ is artinian and noetherian.