Cleardenom does not work correctly over integer and coefficient rings.

[gorzel]

Ok in char 0:

> ring r0 = 0,x,dp;
> cleardenom(-x2+x);
x2-x
> cleardenom(-4x2+2x);
2x2-x
> cleardenom(-4x2+3x);
4x2-3x

If the leading term is -1, then cleardenom should normalize the polynomial. This means that all coefficients have to be multiplied with -1 not only the first one.

See here, small bug over Z, units are +/-1:

>  ring rZ = integer,x,dp; 
>  cleardenom(-x2+x);
-x2+x
>  cleardenom(-4x2+3x);   // it should give 4x2-3x
-4x2+3x

Now over Z_10:

>  ring r10 = (integer,10),x,dp; 
>  cleardenom(3x+6);   // OK
x+2
> 1/number(3);     // 3 is a unit
7
>  cleardenom(3x-6);   // all correct here
x+8
>  cleardenom(-3x-6);
x+2
>  cleardenom(-3x+6);
x+8

How is exactlly cleardenom defined in this case? The following results are OK, but which generator of the units do you choose?

> cleardenom(-4x2+2x);  // *7 , 7== -3
2x2+4x
> cleardenom(4x2+2x);   // * 3
2x2+6x
> cleardenom(4x2+7x); // *3 
2x2+x

But bug now in Z_8:

> ring rZ8 =(integer,8),x,dp;
> cleardenom(-4x2+3x);        // Wrong, by division with -1
4x2+3x                        // it schould give 4x2+5x
>  cleardenom(-x2+x);         // Ok.
x2+7x
> _*(-1)== -x2+x;
1
>  cleardenom(-4x2+2x);       // wrong, it should give 4x2+6x
4x2+2x

[seelisch]

Ok, we now also change sign if the leading coef is negative, i.e. cancel out -1.

General rule is: We compute the largest unit, u, contained in the leading term and then multiply the whole term by the inverse of u. (Internally, u will always be delivered as a positive number. Thus, if the leading coef is negative, we multiply by -1/u, to also change the sign.)

Your examples over Z/8Z are all correct, since -4 = 4 (mod 8).