1. yes. yes. no. (union of subspaces is not a subspace in general) no. (as above) 2. (a) we need to show that if x in C and C' and y in C and C' then any linear combination is also in both. But if x and y in C then any combination is in C since C is linear; and by the same argument any combination is in C', so it is in both. (bi) for the cyclic property we need to show linearity (as above) and also that if x in C and C' then any cyclic shift is in both. But if x in C then any cyclic shift is in C since C is cyclic; and by the same argument x in C' means any cyclic shift is in C', so any cyclic shift is in both. generator: see notes. (bii) Consider w in C+C'. We need to show that its shift is also in C+C'. Put w=a+b with a in C and b in C'. Then the shift a' of a is in C and hence in C+C'. Similarly for shift b' of b in C'. But a'+b', which is thus in C+C', is the shift of w by construction. Done. generator: see notes. (Hint: first consider the code generated by the GCD of g and g', and show it is contained in our code...) 3. the length 99 repetition code. - this has d(C)=99. rate: see notes. (i.e. no indicative answer drafted here yet!) 4. the complete set of quadratics is x^2, x^2 +1, x^2 +x, x^2 +x +1. we substitute x=0,1 in each and look for ones that are never zero. only x^2 +x +1 satisfies this requirement. 5. the complete set is: x^3, x^3 +1, x^3 +x, x^3 +x +1, x^3 +x^2, x^3 +x^2 +1, x^3 +x^2 +x, x^3 +x^2 +x +1. substituting as in 4. above we find x^3 + x +1 x^3 + x^2 +1. what is new with quartics is that a quartic may factorize without a linear factor, so it is not enough to look for quartics that are never zero under substitution, since, for example, (x^2 +x +1)^2 is not irreducible.