Proving that the radical of an ideal is an ideal

Question

I'm trying to solve problem 1.18 in Fulton's 'Algebraic Curves', which is illustrated in the attached image, but I'm having some difficulties understanding the first part.

The ring R is assumed to be commutative and unital, so the first thing that came to mind was to consider the binomial expansion of $(a+b)^{n+m}$, but I don't see why powers of $a$ which are less than n, or powers of $b$ which are less than m should be contained in the ideal.

An explanation as to how I can show that $Rad(I) $ contains $a+b$ would be appreciated, particularly I'd like to know if one can indeed be guaranteed the containment (in I) of the aforementioned terms from the binomial expansion.

I suspect that I may be missing or forgetting some ring theoretic fact.

Answer 1

The fact is that in the binomial expansion you have a sum of products $a^ib^{n+m-i}$ with $0 \le i \le n+m$. So either $i\ge n$ in which case $a^i \in I$ or $n+m-i\ge m$ in which case $b^{n+m-i}$. In both cases $a^ib^{n+m-i} \in I$.

Answer 2

Suppose, for instance, that $m=3$ and that $n=2$. Then$$(a+b)^{m+n}=(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$Then $a^5$, $a^4b$, $a^3b^2$ and $a^2b^3$ all belong to $I$, since $a^2$ does. And $ab^4$ and $b^5$ also belong to $I$, since $b^3$ does. Therefore, $(a+b)^5\in I$.