We aren't going to characterize all the elements of Q′, but we can certainly answer questions 1 and 2 above.
When the linear metric was used, the completion of Q produced R, the real numbers. This brings in lots of new, algebraic elements, like the square root of 2, but R is not algebraically closed, since there is no square root of -1. If the reals are a guide, the answer to (1) will be yes, and the answer to (2) will be no. Indeed, this is the case.
Let q(x) be an irreducible polynomial with integer coefficients. (I didn't want to use p(x); we're already using p for the base prime.)
Let s be a p-adic number that is a solution to q(x). Thus s is a cauchy sequence of rational numbers that produces 0 when substituted into q(x). But how do you replace x with an entire sequence?
When you substitute s for x in q(x), you need to use the operators that were defined for p-adic numbers. As you recall, addition and multiplication are performed term by term. Thus s2 simply squares the terms of s. Similarly, the ith term of sn is the ith term of s, raised to the nth power. Scale these powers of s by the coefficients of q, then add them together. The result is a sequence whose ith term is q(si). Therefore s is a root of q(x) iff q(sn) approaches 0 as n approaches infinity, iff the valuation of q(sn) approaches infinity as n approaches infinity. We are looking for a sequence of rational numbers, such that q(sn) is divisible by ever higher powers of p.
Let's start the sequence by assuming there is an integer c such that q(c) is divisible by p. If q(c) = 0 then c is a root of q, and we are done. Otherwise let q(c) = gpj. Now c is the first term in our sequence, s0 = c, and the valuation of q(c) equals j.
Next, evaluate q(c+pj) mod pj+1. Remember that q(c) = gpj, and move from c to c+pj. Start with the linear term. Multiply the linear coefficient by pj, and add this to gpj. Then move to the squared term. Multiply its coefficient by 2cpj, and add this to the total. Expand (c+pj)3 using the binomial theorem, and bring in 3c2pj times the coefficient on the cubed term. Add 4c3pj times the coefficient on the fourth term, and so on. This gives the value of q(c+pj) mod pj+1.
Add up the linear coefficient, plus 2c times the coefficient on the squared term, plus 3c2 times the coefficient on the cubed term, and so on up to ncn-1 times the leading coefficient, and call this sum u. We are only interested in u mod p.
Assume u is nonzero, and let w = -g/u. Evaluate q(c+wpj). Now w is multiplied by u, then added to g. This is 0 mod p, hence q(c+wpj) is divisible by pj+1. The valuation has increased.
Set s1 = c+wpj, and repeat the above procedure to find s2, s3, and so on. Thus s is a root of q(x).
The integer p2+1 has no square root in Q, but it always has a square root in Q′. Reduce mod p, and p2+1 becomes 1, and 1 always has a square root mod p, namely 1. Set s1 = 1 to start the sequence. Also, the sum of the coefficients times their degree, in q(x) = x2-(p2+1), is 2, which is nonzero mod p, at least when p is odd. The induction step is valid, and the process described above builds the entire sequence s.
If p = 2, Q′ contains the cube root of 3, but Q does not. Again, 1 seeds the sequence, and the sum of the coefficients times degree, in x3-3, is 3, which is 1 mod 2.
We have answered question (1) above. Q is not complete with respect to the p-adic topology. Its completion Q′ brings in additional elements, including quite a few algebraic elements that were not part of Q. But don't confuse the square root of 2 in the 7-adic topology with 1.4142135…, which is the square root of 2 in the linear topology. The numbers are both sequences, or series if you prefer, but they bear no relationship to each other, because the metrics are different. In fact the p-adic square root is a series of integers that grow larger and larger (in the traditional sense). However, these integers, when squared, do indeed approach 2 under the p-adic metric. This is so weird, maybe we should derive the first few terms.
since 32 = 2 mod 7, set s0 = c = 3. Now q(3) = 7, or 1×7 if you prefer. The value of u is 6, so our first instance of w is -1/6, or 1. The next term in the sequence is therefore 3+1×7, or 10.
Evaluate q(10) and get 98, which is 2×72. Since -1/u is conveniently 1, just use the coefficient 2 as is. Therefore s2 = 10+2*49 = 108. Square this and subtract 2 to get 34×73, whence s3 = 108+6*343 = 2166. The next value of q is 1954×74, hence s4 = 2166+1×2401 = 4567. Continue this process to build sqrt(2).
Let's see if we can answer question (2). This time we are looking for a q(x) that has no solution in the p-adic numbers.
Let q(x,y) be the balanced polynomial associated with q(x). For instance, if q(x) = 3x3+5x2+7x+9, we obtain the following.
q(x,y) = 3x3 + 5x2y + 7xy2 + 9y3
If q has degree n, and q(a,b) = e, where a b and e are integers, divide through by bn to show q(a/b) = e/bn. Conversely, assume q(a/b) is some rational number that we will call e/bn. Multiply through by bn, whence q(a,b) = e, and e is an integer. This relationship helps us evaluate q(x) for rational values of x.
Suppose there is no integer c such that q(c) = 0 mod p.
Let a/b be a rational number, a term in the sequence s. Since q(s) approaches 0 in our p-adic metric, the valuation of q(s) approaches infinity. Select a high valuation v, and find an entry in the sequence s, a rational number a/b, such that q(a/b) has valuation v.
If q(a/b) = e/bn, and b is divisible by pj, then e is divisible by pnj+v. Start with the simplest case, j = 0. Thus b is not divisible by p. Switch to the symmetric form q(a,b) = e. Reduce mod p and divide through by bn. This takes us back where we started from, q(a/b) = e/bn, but a/b is no longer a rational number; it is an integer mod p. Call it c. Now q(c) = 0 mod p, and that is a contradiction. Therefore b is divisible by p, and a is not. For most of the sequence, si has a negative valuation.
Return to the symmetric form q(a,b) = e. The valuation of e is at least nj+v. What is the valuation of the first term of q(a,b)? Since p does not divide a, it is simply the valuation of the lead coefficient. Let this valuation be k. In other words, the lead coefficient of q(x) is divisible by pk. When j exceeds k, the valuation of q(a,b) is precisely k. Remember that v can be as high as we like; certainly higher than k. To produce a valuation of v, k must be at least as large as j. Beyond some point in the sequence, all the terms of s have valuation between -k and -1 inclusive.
In fact the valuation becomes constant. If two terms have different valuations their difference has the lesser valuation. This is negative, and nowhere near v. Since s is cauchy, the valuation becomes fixed at -j. Once again, -j is trapped between -k and -1.
If k = 0 there is no solution. The terms of s cannot have valuation between -0 and -1. Let's look at a couple of examples.
When p = 7, and q = x2+1, there is no integer c such that c2 = -1 mod 7, and the lead coefficient is certainly not divisible by 7, hence x2+1 has no solution in the 7-adic numbers. There is no square root of -1 in this field. Q′ is not algebraically closed, nor does it contain the algebraic closure of Q. The answer to (2) is no, as we expected.
This generalizes to any prime p. Let w be an integer that is not a square mod p, and w does not have a square root in the p-adic numbers. When p = 2, there is no solution to x2+x+1 in the 2-adic numbers.
In summary, the p-adic numbers bring in some, but not all, of the algebraic elements of Q.