Looking for an exemple of $ \mathbb{Z} $ - linear application $ f : \mathbb{Z}^n \to \mathbb{Z}^m $ satisfying the following conditions :

Question

I'm looking for :

• An exemple of $ \mathbb{Z} $ - linear application $ f : \mathbb{Z}^{ (\mathbb{N}) } \to \mathbb{Z}^{ (\mathbb{N}) } $ satisfying the following condition :

$$ \exists y \in \mathbb{Z}^{(\mathbb{N})} \ \ \forall p \in \mathbb{Z} \backslash \{ 0 \} \ \ \forall x \in \mathbb{Z}^{(\mathbb{N})} \ : \ py \neq f(x)$$

I'm also looking for :

• Another exemple of $ \mathbb{Z} $ - linear application $ f : \mathbb{Z}^n \to \mathbb{Z}^m $ such that : $ n \geq m $ satisfying the following condition :

$$ \exists y \in \mathbb{Z}^{m} \ \ \forall p \in \mathbb{Z} \backslash \{ 0 \} \ \ \forall x \in \mathbb{Z}^{n} \ : \ py \neq f(x)$$

Thanks in advance for your help.

Answer 1

For the first question (under the assumption $p \neq 0$, as otherwise Anurag's comment would kick in) an answer would be

$$f(\{z_i\}) = \{iz_i\},$$

with $y$ chosen as $y_i=1, i=1,2,\ldots$. For any $p \neq 0$, $(py)_i=p, i=1,2,\ldots$ but at indices $n > |p|$ we have for any $x$ either $(f(x))_n = 0$, or $|(f(x))_n| \ge n > |p|$, so $(f(x))_n = p$ is impossible for those $n$.

Answer 2

With the modified statement. Just choose $f$ to be the zero function. Then $f$ is linear for sure. And it's output cannot be any non-zero vector, so there exists a $y$ such that $f(x) \neq py$ for all non-zero integers $p$.

For second part: Let $f:\mathbb{Z}^3 \longrightarrow \mathbb{Z}^2$ be given by $f(x,y,z)=(2x,0)$. Then this is linear. But $f(x,y,z) \neq p(1,1)$ for all $(x,y,z) \in \mathbb{Z}^3$