# How do you calculate the Euler factors of the imprimitive symmetric square at primes with bad reduction?

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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Hard to parse the notation. To recap: $D_r(X)$ and $\rho_r′$ correspond to taking $I_r$ invariants on $H_1^l(E)$ and then taking the ${\rm Sym}^2$, while the more standard $\rho_r$ first takes ${\rm Sym}$ and then takes $I_r$ invariants. So I think the answer is: when $E$ is additive reduction, then $H^1_l(E)^{I_r}$ is trivial, and so is the symmetric square of it. When $E$ is multiplicative reduction, then $H_1^l(E)^{I_r}$ is a 1-dim subspace, and in fact is the same as with $\rho_r$ if I am not mistaken (it has degree 1, and divides the degree 1 Euler factor with $\rho_r'$, so must be same) –  Junkie Jun 10 '11 at 7:48
ah yes, thank you. –  Max Flander Jun 10 '11 at 13:01

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.