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From this answer and this answer, I can understand how to proof this statement. But, i got confused at

Let $f(x)=a_0+a_1x+a_2x^2+⋯+a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+⋯+b_mx^m$ be two elements of $R[x]$. Let $0≤i≤n$ and $0≤j≤m$ such that $i$ and $j$ are the smallest numbers such that $a_i≠0$ and $b_j≠0$.

With $f=a_0+a_1x+…$ and $g=b_0+b_1x+…$ let $a_i$ and $b_j$ be respectively the coefficients of the smallest non-zero terms in $f$ and $g$.

Why are $i$ and $j$ stated as the smallest non-zero terms? For example, given $f = 6 + 3x + 5x^2 + ...$, that means we pick $i = 0$ right?

If we pick $i$ and $j$ like this, we can always prove that the result (coef. of $x^{i+j}$) is non-zero. But, latter in the first answer proof, we conclude that $\forall s > i, a_s = 0$. This is contradiction for our $i = 0$ before, because there is $i = 1$ such that $a_1 \neq 0$.

Can $i$ and $j$ picked as the largest non-zero terms?

P.S. I know there is another way to prove this theorem, I just confused with what I asked.

ausfear
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    The correct conditions are that $,a_i,$ is the lead coef of $f\neq 0$, i.e. $,a_i\neq 0,$ and $,a_k = 0,$ for all $,k>i.,$ Similarly $b_j$ is the lead coef of $g$. Likely it is stated in this unusual way to apply to the proof of Gauss's Lemma, as I explain in my answer there. In cases like this you should first pose questions in comments on the prior answer(s) to avoid duplication (and help improve prior answers). – Bill Dubuque May 20 '22 at 01:44
  • @BillDubuque Actually I want to comment on those answers, but my account can't comment on others' post yet, so I decide to ask a new question. However, thanks for the answer. – ausfear May 20 '22 at 01:53
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    Now you should be able to comment. – Bill Dubuque May 20 '22 at 01:55

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