The reference for this question is Coates and Schmidt, Iwasawa theory for the symmetric square.


Let $G = \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}))$ and let $D\_r \supseteq I\_r$ be a decomposition group and inertia group at $r$. Let $E/\mathbb{Q}$ be an elliptic curve, and for a prime $\ell \neq r$, consider the representations $\rho\_r : G\to \textrm{Sym}^2(H^1\_\ell(E))^{I\_r}$ and $\rho\_r' : G\to \textrm{Sym}^2(H^1\_\ell(E)^{I\_r})$. 


The primitive symmetric square of $E$ is the $L$-series defined by the Euler factors


$\mathcal D\_r (X) = \textrm{det}(1-\rho\_\ell(\textrm{Frob}\_r^{-1})X)$ 


and the imprimitive symmetric square of $E$ is the $L$-series defined by the Euler factors $D\_r (X) = \textrm{det}(1-\rho\_\ell'(\textrm{Frob}\_r^{-1})X).$


Since $\rho'\_r$ is a submodule of $\rho\_r$, we have that $D\_r(X) | \mathcal{D}\_r(X)$ 


for all $r$. Furthermore, if $r$ is a prime where $E$ has good reduction, since $H^1\_\ell(E)$ is unramified at $r$ we have $D\_r(X) = \mathcal{D}\_r(X)$ .


If $E$ has bad multiplicitive reduction at $r$, then by calculations in Coates and Schmidt, $\mathcal{D}\_r(X) = 1-X$, and if $E$ has bad additive reduction at $r$, then $\mathcal{D}\_r(X)$ is either equal to $(1-\alpha\_r^2X)(1-\beta\_r^2X)(1-rX)$, $1+rX$, $1-rX$ or $1$ depending on the image of the inertia group $I\_r$, and I can calculate this using Mark Watkins's 'sympow' computer program.


My question is:



> 
> How do I find the imprimitive Euler factors $D\_r(X)$ at primes where $E$ has bad reduction?
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> 
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How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction?
The factor $D\_r$ is easy to compute (much easier than $\mathcal{D}\_r$). Basically, you just need to find the eigenvalues $\lambda\_i$ of Frobenius on $H^1\_\ell(E)^{I\_r}$ (i.e. the reciprocal roots of the local L-factor of E itself), and then the eigenvalues of Frobenius on the symmetric square of that are the pairwise products $\lambda\_i \lambda\_j$. This gives you the reciprocal roots of $D\_r$.


1. If $r$ is a good prime, the eigenvalues of Frob on $H^1\_\ell(E) = H^1\_\ell(E)^{I\_r}$ are $\alpha\_r$ and $\beta\_r$, so on the symm square you get $\alpha\_r^2$, $\beta\_r^2$ and $\alpha\_r \beta\_r = r$. So have $D\_r(X) = \mathcal{D}\_r(X) = (1 - \alpha\_r^2 X)(1 - \beta\_r^2 X)(1 - r X)$.
2. If $r$ is a bad multiplicative prime, then $H^1\_\ell(E)^{I\_r}$ is 1-dimensional and the only eigenvalue of Frob is $\pm 1$ (depending whether the reduction is split or non-split). Either way the only eigenvalue on the symmetric square is $+1$, and $D\_r(X) = (1 - X)$.
3. If $r$ is a bad additive prime, then $H^1\_\ell(E)^{I\_r} = 0$ and hence $D\_r(X)$ is identically 1.


The calculation of $\mathcal{D}\_r$ is much harder, because you need to consider the case where the image of inertia is some finite subgroup of $\mathrm{GL}\_2(\mathbb{Z}\_\ell)$ which has zero invariants but preserves some bilinear form. 


