Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 3

An ideal $I$ of a commutative ring is an additive subgroup such that $rx\in I$ for every $r\in R$ and $x\in I$. A prime ideal is a proper ideal $A\ne R$ such that $xy\in A$ implies $x\in A$ or $y\in A$.

If $(x+A)(y+A)=0+A$ in the quotient ring, then $xy\in A$. Primality implies $x+A=0+A$ or $y+A=0+A$. Therefore there are no two nonzero cosets whose product is zero:Properness also ensures its identity coset is not zero, so this quotient is an integral domain.

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.

A coset is a nilpotent element of the quotient exactly when some positive power is zero:If $J=\sqrt{J}$, this forces every nilpotent coset to equal zero. Conversely, if there is no nonzero nilpotent coset, every $x\in \sqrt{J}$ has $x+J=0$, hence $x\in J$. Since $J\subseteq \sqrt{J}$ always, equality follows. ThereforeIn standard terms, an ideal is radical exactly when its quotient is a reduced ring.

Certainly $A\subseteq \sqrt{A}$, by taking exponent one. If $x\in \sqrt{A}$, choose $n\ge 1$ with $x^n\in A$. Primality applied to $x\cdot x^{n-1}$ gives $x\in A$ or $x^{n-1}\in A$. In the latter case repeat; induction reduces to $x\in A$. Thus $\sqrt{A}\subseteq A$ as well, provingEvery prime ideal is therefore a radical ideal.