Isomorphisms of Tensor products of Sylow p-subgroups

Let A be a finite abelian group of order n and let p^k be the largest power of the prime p dividing n. Prove that (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A is isomorphic to the Sylow p-subgroup of A.

Proof:
We know that, if M is a finite abelian group of order a = p_1^{\alpha _1}p_2^{\alpha _2} \cdots p_k^{\alpha _k} then considered as a \mathbb{Z}-module, M is annihilated by (a), the p_i-primary component of M is isomorphic to the direct product of its Sylow subgroups.
Here, A is a finite abelian group of order n. As a \mathbb{Z}-module, A is the direct sum of its Sylow p-subgroups, say A = \oplus _{q\in Q}A_q where Q is the finite set of primes dividing n. Tensor product commute with finite direct sums, so (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A \cong (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}(\oplus_{q\in Q} A)\cong \oplus _{q\in Q}(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q as \mathbb{Z}-homomorphism.
We show if p\neq q, then (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q = 0. Since p and q are relatively prime (in this case) in the PID \mathbb{Z}, then there are a,\, b \in \mathbb{Z} s.t. ap^k + bq^l =1. Let \overline{x}\otimes y be a simple tensor in (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q. Then, \overline{x}\otimes y = \overline{x}\cdot 1 \otimes y = \overline{x}(ap^k+bq^l)\otimes y = \overline{x}ap^k\otimes y + \overline{x}bq^l\otimes y = \overline{xp^k}a\otimes y + \overline{x}b\otimes q^ly = 0\otimes y + \overline{x}b\otimes 0 = 0. Hence, (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q = 0.
Now we show (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p. Here, every simple tensor in (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p can be written as \overline{1}\otimes x. Then any tensor, \overline{a}\otimes x = \overline{1}a\otimes x = \overline{1}\otimes ax. Now define \phi :(\mathbb{Z}/p^k\mathbb{Z})\times A_p \rightarrow A_p by \phi (\overline{a},\,x)=ax. Here, if p^k divides (a-b) then (a-b)x =0 \Rightarrow ax = bx. Hence \phi is well-defined. It is easy to see that \phi is a \mathbb{Z}-balanced map, so \phi induces a group homomorphism \Phi : (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \rightarrow A_p and \iota is the necessary natural inclusion and \Phi \circ \iota = \phi.

Let (\overline{a},x)\in (\mathbb{Z}/p^k\mathbb{Z})\times_{\mathbb{Z}}A_p then (\Phi \circ \iota)(\overline{a},x)=\phi(\overline{a},x) \Rightarrow \Phi (\overline{a}\otimes x) = ax.
It is easy to see that \Phi is injective. Also, \Phi is surjective, since for each x\in A_p, \Phi(\overline{1}\otimes x)=x. Thus \Phi is an isomorphism, i.e. (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p.

Hence, (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A \cong (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p.

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Ideal generated by irreducible polynomial and its Quotient Ring

Let F be a field and x be an indeterminate over F. Let f(x) \in F[x].Prove that F[x]/(f(x)) is a field if and only of f(x) is irreducible.

Proof:
We know that for any ring R, R/I is a field if and only if I is a maximal ideal of R. To use it, we show that (f(x)) is maximal if and only if f(x) is irreducible.
"\Leftarrow" Let I = (f(x)) be maximal. We show 0\neq f(x) is irreducible. If possible, let f(x) = g(x)h(x) where the degree of g(x) and h(x) is at least one, for some g(x),\, h(x)\in F[x]. Here, g(x)|f(x) and h(x)|f(x) so the ideals J_1=(g(x)) and J_2=(h(x)) are proper and also contains I, which is a contradiction since I is maximal. So the polynomial is irreducible.
"\Rightarrow" Let f(x) be irreducible. We show I=f(x) is maximal. If possible, let I\subsetneq J for an ideal J of F[x]. Since F is a field so F[x] is a PID, so every ideal of f[x] is principal. Then for a polynomial g(x)\in F[x] is principal. Then for a polynomial g(x)\in F[x], J = (g(x)).
Since f(x)\in I \subsetneq J, so for f(x) \in J there is some q(x)\in F[x] s.t. f(x)=q(x)g(x). Since f(x) is irreducible, so either q(x) or g(x) is of degree zero. If deg\, f(x)=0 then g(x) is a unit in F[x]. Here, g(x)\in J i.e. J contains a unit, so J = F[x]. If deg\, q(x)=0 then f(x)=cg(x), for q(x)=c, constant. Then g(x)\in (f(x))=I i.e. J = (g(x))\subsetneq I i.e. J\subsetneq I. Hence, I = J. Which suggests that I=(f(x)) is maximal.
"\Leftarrow" Now, let F[x]/(f(x)) is a field, so the ideal (f(x)) is maximal. We have that if (f(x)) is maximal then f(x) is irreducible.
"\Rightarrow" Let f(x) is irreducible then (f(x)) is maximal so F[x]/(f(x)) is a field.