If we have encryptions of additive and multiplicative identities in the corpus of cipher text of a deterministic fully homomorphic encryption (FHE) scheme, I guess we can break it.
If the FHE scheme is deterministic and works over integers and if the corpus of cipher text has $enc(1),enc(0)$ where $1$ being multiplicative identity and $0$ is additive identity for integers, I am thinking it is easy to break such scheme.
First we can verify if there any below inferences in the cipher text
if $eval(enc(x) + enc(y)) = enc(x)$ this means $enc(y)$ is encryption of $0$.
if $eval(enc(x) * enc(z)) = enc(z)$ this means $enc(z)$ is encryption of $1$.
if $eval(enc(x) + enc(a)) = enc(0)$ this means $a$ is additive inverse i.e $-x$.
if $eval(enc(x) * enc(b)) = enc(1)$ this means $b$ is multiplicative inverse i.e $1/x$.
if $eval(enc(x) + enc(c)) = enc(1)$ this means $x+c = 1$
Now it is easy to identify all the numbers without decrypting them.
if $eval(enc(w)+ n\times (enc(1)) = enc(0)$ this means $w= -n$.
Is this understanding correct ?