The shell quoting rules are different between Windows and Linux/UNIX. Within double quotes (") the $ sign is significant to the Linux/UNIX shell. Try:
>perl -Mntheory=:all -nE 'chomp; say is_prime($_);' input.txt
The argument for the -e flag is surrounded with " on Windows, but with ' on Linux/UNIX. Quoting rules therefore are different in the body of the script provided via -e, which led iirc to the q() family of operators (see perlop)
update: the _ identifier is set in the shell to the last argument of the previous command, so $_ is set to input.txt which is interpolated into the program string passed to perl via -e as the expression of concatenation of two barewords:
qwurx [shmem] ~> perl -E "say '$_'" foo.bar
qwurx [shmem] ~> perl -E "say '$_'" foo.bar
foo.bar
qwurx [shmem] ~> perl -E "say $_" foo.bar
foobar
This is a dangerous thing, if the last argument of the previous line happens to contain metacharacters, e.g.
qwurx [shmem] ~> echo '`ls -l /dev/null`'
`ls -l /dev/null`
qwurx [shmem] ~> perl -E "say $_"
crw-rw-rw- 1 root root 1, 3 Nov 27 11:02 /dev/null
qwurx [shmem] ~> perl -MO=Deparse -E "say $_"
use feature 'current_sub', 'evalbytes', 'fc', 'say', 'state', 'switch'
+, 'unicode_strings', 'unicode_eval';
say say(say(`ls -l /dev/null`));
-e syntax OK
so don't do that.
perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'
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