Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 3 / Solution

A maximal ideal is a proper ideal $I$ of $R$ such that an ideal $J$ with $I\subseteq J\subseteq R$ is either $I$ or $R$. Suppose first that $I$ is a maximal ideal. If $a\notin I$, then $I+(a)=R$, so $1=i+ra$ for some $i\in I$ and $r\in R$. In the quotient ring, $(r+I)(a+I)=1+I$. Every nonzero element of $R/I$ therefore has a multiplicative inverse, so $R/I$ is a field. Conversely, if $R/I$ is a field and $J$ properly contains $I$, choose $a\in J\setminus I$. The element $a+I$ has a multiplicative inverse $r+I$, whence $1-ra\in I\subset J$, giving $1\in J$ and $J=R$. ThusThe squares in ${\mathbb{F}}_7$ are $0,1,2,4$, which do not include $-1=6$. Hence $X^2+1$ has no root in ${\mathbb{F}}_7$. A polynomial of degree two or three over a field is an irreducible polynomial exactly when it has no root: any proper polynomial factorization of such a degree has a factor of degree one. The polynomial ring over a field is a Euclidean domain, so the ideal generated by an irreducible polynomial is a maximal ideal. It follows thatis a field. Every coset has a unique representative $a+bX$, by polynomial division, so $|F|=7^2=49$.

Let $j$ be the coset of $X$ in this field. Since $j^2=-1$, the mapis a ring homomorphism. It is surjective by the representative description. Since $1,j$ are linearly independent over ${\mathbb{F}}_7$, its kernel of a ring homomorphism isThere is also a field with $343$ elements. The cubes in ${\mathbb{F}}_7$ are $0,1,-1$, so $X^3-2$ has no root and is an irreducible polynomial. Thereforeis a field whose unique representatives have degree below three. There are $7^3=343$ such representatives.