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Grothendieck group of coherent sheaves is not a ring?

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My question is motivated by the fact that the Grothendieck group $K^0(X)$ of vector bundles on $X$ can be given a ring structure via the tensor product. But it seems to me that the Grothendieck group of coherent sheaves $K_0(X)$ has no such structure. Why?

Let $X$ be any scheme. Denote by $K_0(X)$ the Grothendieck group of coherent sheaves on $X$, defined as the quotient of the free abelian group $G_{\text{coh}}$ generated by formal symbols $[\mathscr F]$, where $\mathscr F$ is a coherent sheaf on $X$, by the relations $[\mathscr F] = [\mathscr F_1] + [\mathscr F_2]$ whenever there is a short exact sequence $0\to \mathscr F_1 \to \mathscr F\to \mathscr F_2 \to 0$.

It seems that the tensor product defines a ring structure on $G_{\text{coh}}$, so I assume that the subgroup generated by $[\mathscr F] - [\mathscr F_1] - [\mathscr F_2]$ is not an ideal in $G_{\text{coh}}$. Is there a concrete example of this?

Replacing every occurrence of "coherent sheaf" by "vector bundle", we obtain the Grothendieck group $K^0(X)$ of vector bundles on $X$. My understanding is that tensor product on $G_{\text{vb}}$ descends to a ring structure on $K^0(X)$, i.e. the subgroup of $G_{\text{vb}}$ generated by $[E] - [E_1] - [E_2]$is an ideal of the ring $G_{\text{vb}}$.

Is there a philosophical reason why this should hold for $K^0$ but not $K_0$?


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