Doubts in proof of FTA (Fundamental Theorem of Arithmetic) I have issues in understanding the proof of the F.T.A. by Wallace in the book titled: Groups, Rings and Fields, on page #66. It is stated that : "Then $p_1$ divides $q_1q_2...q_n$". I feel $q_n$is a typo, and should be $q_s$. (ii) After finding that with suitable reordering, $p_1 = q_1$, the prime $p_r$ is replaced with $q_r$, and $q_1$ with $p_1$. So, it means the replacement of $p_r$ is done with $q_r$, for the same reason as shown for replacing $q_1$ with $p_1$. (iii) It is stated that : "By our induction assumption we have $r=s$". But, it was only 'supposed' by the statement: "Suppose now that $n=p_1p_2...p_r = q_1q_2...q_s$", that the two prime factorization are equal to $n$, nothing more. Also, to vindicate my conjecture, in the issue (ii) it is shown in the book that $q_r$ replaces $p_r$; rather than $q_s$ replacing $p_r$.

The issues are : (i) It is stated that : "Then p1 divides q1q2...qn". feel qn is a typo, and should be qr.

Note: You are correct. This is a Typo.

(ii) After finding that with suitable reordering, p1=q1, the prime pr is replaced with qr, and q1 with p1. So, it means the replacement of pr is done with qr, for the same reason as shown for replacing q1 with p1.

Note: At this step we just one to eliminate one of the primes from each side.

(iii) It is stated that : "By our induction hypothesis we have r=s". But, it was only 'supposed' by the statement: "Suppose now that n=p1p2...pr=q1q2...qs", that the two prime factorization are equal to n, nothing more.

Note: The induction hypotheses clearly indicates that if n=p1p2...pr=q1q2...qs then r=s, therefore there is no flaws in the proof.