Commutativity of Ring with polynomial elements

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You need to prove that polynomials satisfy basic axioms of ring for addition and multiplication.

Let $P(x)=\sum_{i=0}^na_ix^i, \:Q(x)=\sum_{i=0}^mb_ix^i$.Then $$ P(x)+Q(x)=\sum_{i=0}^n(a_i+b_i)x^i=\sum_{i=0}^n(b_i+a_i)x^i=Q(x)+P(x) $$ This proves polynomials is commutative for addition. $$ P(x)Q(x)=\sum_{i=0}^n\sum_{j=0}^i(a_jb_{i-j})x^i=\sum_{i=0}^n\sum_{j=0}^i(b_{i-j}a_j)x^i=Q(x)P(x) $$ This proves polynomials is commutative for multiplication.

So I found this proof and I don't understand where the $m$ went from $Q(x)$. Why does the $m$ disappear when we add $P(x)$ and $Q(x)$? Was this an error?

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Without loss of generality you can assume that both polynomials have the same degree, as you can always extend it by zero coefficients. This is common practice, so I think the author simply forgot to mention it (which is a fauxpas but happens)