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2 Flagged up my misunderstanding of the question!; added 2 characters in body

[EDIT: I misunderstood the question -- I thought the poster wanted to know why all crystalline characters are unramified twists of powers of cyclotomic when $L$ or $K$ is $\mathbb{Q}_p$, and wrote out a detailed proof. I realise now that Laurent references answers the poster already knew this question in fine style; but I found it somewhat hard wanted to read, since it's proving something vastly more understand the classification in the general , so (mostly for my own benefit!) here's an alternative argumentcase. I thought I'd leave this post here anyway in case anyone finds it useful.]

Step 1: Consider the space $\mathbb{D}_{\mathrm{dR}}(\chi)$. This is a free rank 1 module over $K \otimes_{\mathbb{Q}_p} L$, and the steps in the Hodge filtration are $K \otimes_{\mathbb{Q}_p} L$-submodules. By assumption, $K \otimes_{\mathbb{Q}_p} L$ is a field, there is only one Hodge-Tate weight. By twisting by a power of the cyclotomic character, we can assume without loss of generality that the Hodge-Tate weight is 0.

Step 2: Now consider the space $\mathbb{D}_{\mathrm{cris}}(\chi)$. This is a free rank 1 module over $L K_0$, with a Frobenius that is L-linear and $K_0$-semilinear. (Here $K_0$ is the maximal unramified subfield of $K$). If $[K_0 : K] = d$, then $\varphi^d$ is linear, so acts as multiplication by a scalar $\mu \in L$.

Step 3: Consider the unramified character $G_{K_0} \to L$ mapping geometric Frobenius to $\mu^{-1}$. Tensoring with this character, we may assume that $\mu = 1$.

Step 4: Now choose a basis of $\mathbb{D}_{\mathrm{cris}}(\chi)$ over $LK_0$, and suppose that (with respect to this basis) $\varphi$ acts as multiplication by $\alpha$. The condition that $\varphi^d = 1$ translates to $\alpha \varphi(\alpha) \dots \varphi^{d-1}(\alpha) = 1$. If $K = \mathbb{Q}_p$, then this is just $\alpha = 1$. If $K \ne \mathbb{Q}_p$, then $L = \mathbb{Q}_p$, and $\alpha \in K_0^\times$ satisfies $N_{K_0 / \mathbb{Q}_p} \alpha = 1$. Hence $\alpha = \varphi(x) / x$ for some $x \in K_0^\times$, by Hilbert 90; and after changing basis by $x$ we again have $\alpha = 1$.

So we've shown that $\mathbb{D}_{\mathrm{cris}}(\chi)$ is isomorphic (as a filtered $\varphi$-module) to $\mathbb{D}_{\mathrm{cris}}$ of the trivial character, i.e. $\chi$ is trivial (and hence crystalline!). QED.

1

The paper by Conrad that Laurent references answers this question in fine style; but I found it somewhat hard to read, since it's proving something vastly more general, so (mostly for my own benefit!) here's an alternative argument.

Step 1: Consider the space $\mathbb{D}_{\mathrm{dR}}(\chi)$. This is a free rank 1 module over $K \otimes_{\mathbb{Q}_p} L$, and the steps in the Hodge filtration are $K \otimes_{\mathbb{Q}_p} L$-submodules. By assumption, $K \otimes_{\mathbb{Q}_p} L$ is a field, there is only one Hodge-Tate weight. By twisting by a power of the cyclotomic character, we can assume without loss of generality that the Hodge-Tate weight is 0.

Step 2: Now consider the space $\mathbb{D}_{\mathrm{cris}}(\chi)$. This is a free rank 1 module over $L K_0$, with a Frobenius that is L-linear and $K_0$-semilinear. (Here $K_0$ is the maximal unramified subfield of $K$). If $[K_0 : K] = d$, then $\varphi^d$ is linear, so acts as multiplication by a scalar $\mu \in L$.

Step 3: Consider the unramified character $G_{K_0} \to L$ mapping geometric Frobenius to $\mu^{-1}$. Tensoring with this character, we may assume that $\mu = 1$.

Step 4: Now choose a basis of $\mathbb{D}_{\mathrm{cris}}(\chi)$ over $LK_0$, and suppose that (with respect to this basis) $\varphi$ acts as multiplication by $\alpha$. The condition that $\varphi^d = 1$ translates to $\alpha \varphi(\alpha) \dots \varphi^{d-1}(\alpha) = 1$. If $K = \mathbb{Q}_p$, then this is just $\alpha = 1$. If $K \ne \mathbb{Q}_p$, then $L = \mathbb{Q}_p$, and $\alpha \in K_0^\times$ satisfies $N_{K_0 / \mathbb{Q}_p} \alpha = 1$. Hence $\alpha = \varphi(x) / x$ for some $x \in K_0^\times$, by Hilbert 90; and after changing basis by $x$ we again have $\alpha = 1$.

So we've shown that $\mathbb{D}_{\mathrm{cris}}(\chi)$ is isomorphic (as a filtered $\varphi$-module) to $\mathbb{D}_{\mathrm{cris}}$ of the trivial character, i.e. $\chi$ is trivial (and hence crystalline!). QED.