 Solution

The set $\sqrt{I}$ contains zero. If $x^m\in I$, then $(-x{)}^m=(-1{)}^m{x}^m\in I$, giving closure under additive inverses; also $(rx{)}^m=r^m{x}^m\in I$ for every $r\in R$.

To prove closure under addition, let $x^m,y^n\in I$. Expand $(x+y{)}^{m+n-1}$ by the binomial theorem. In every monomial $x^j{y}^{m+n-1-j}$, either $j\ge m$ or $m+n-1-j\ge n$. Thus each term contains a factor $x^m$ or $y^n$ and belongs to $I$. Hence $(x+y{)}^{m+n-1}\in I$, so $x+y\in \sqrt{I}$. The additive subgroup and absorption properties are proved, givingThis ideal is the radical of an ideal. The finite exponent for a sum uses commutativity of the ring.