Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 3

A prime ideal is a proper ideal $I\ne R$ such that $ab\in I$ implies $a\in I$ or $b\in I$. A maximal ideal is a proper ideal $I\ne R$ such that there is no ideal strictly between $I$ and $R$.

Every ideal of the integers has the form $(n)=n\mathbb{Z}$ for a unique nonnegative integer $n$. Its prime ideals and maximal ideals are respectivelyThe zero ideal is a prime ideal because the integers form an integral domain. For a prime number $p$, divisibility of $ab$ by $p$ forces divisibility of one factor, so $(p)$ is a prime ideal. Also any ideal $(d)$ containing $(p)$ has $d\mid p$, so it equals $(p)$ or $(1)$; thus $(p)$ is a maximal ideal.

If $n>1$ is composite, write $n=ab$ with $1<a,b<n$. Then $ab\in (n)$ while neither factor belongs to $(n)$, so $(n)$ is not a prime ideal. A prime number $p\mid n$ gives $(n)⊊(p)⊊\mathbb{Z}$, so $(n)$ is not a maximal ideal either. The zero ideal is not maximal since $(0)⊊(2)⊊\mathbb{Z}$, and $(1)=\mathbb{Z}$ is not proper.

If the quotient ring $R/I$ is an integral domain, then $I\ne R$, because an integral domain has $1\ne 0$. For $ab\in I$,There are no nonzero zero divisors in an integral domain, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving $I$ is a prime ideal.

If $R/I$ is a field, again $I$ is proper. Let $J$ be an ideal properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a multiplicative inverse $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore $I$ is a maximal ideal. These arguments establish the relevant prime ideal quotient criterion and maximal ideal quotient criterion.

Set $X=0$ to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The first isomorphism theorem for rings therefore givesThe polynomial ring $\mathbb{Q}[Y]$ is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: $Y$ has no polynomial multiplicative inverse, since a nonzero product $YQ(Y)$ has positive degree of a polynomial. Consequently $(X)$ is prime and is not maximal. Concretely, $(X)⊊(X,Y)⊊\mathbb{Q}[X,Y]$ is a strict chain of ideals.