Is it decidable whether one language is polytime reducible to another?
I'm going to assume that your question is whether the language $$L = \{\langle M_1\rangle; \langle M_2\rangle\mid L(M_1)\leq_\mathrm{p} L(M_2)\}$$ is decidable, where $\langle M\rangle$ is the description of Turing machine&nbsp;$M$ and $\leq_\mathrm{p}$ denotes polynomial-time many-one reducibility.

This language is undecidable.

The only language that is many-one reducible to&nbsp;$\Sigma^*$ is $\Sigma^*$ itself.  This is because a many-one reduction from $X$ to&nbsp;$\Sigma^*$ must, by definition, map all "no" instances of&nbsp;$X$ to "no" instances of&nbsp;$\Sigma^*$. However, $\Sigma^*$&nbsp; has no "no" instances, so the reduction can only exist if $X$&nbsp;also has no "no" instances, i.e., if $X=\Sigma^*$.

So, now let $Y$ be any Turing machine that accepts every input.  The language $H=\{\langle M\rangle\mid \langle M\rangle;\langle Y\rangle\in L\}$ is clearly reducible to&nbsp;$L$. But $H$&nbsp;is the language of Turing machines that accept every input, which is undecidable.