Past exam of the rings and modules course of the University of Oxford / 2023 / 1 / a

Multiplication in a direct product of rings is coordinatewise. Thus $(r_1,\dots ,r_n)$ has a two-sided inverse exactly when each $r_j$ has a two-sided inverse in $R_j$, and thenConsequently its unit group is the direct product of groupsThe identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

Write $T={\prod }_{j=1}^n{R}_j$, and let ${\rho }_j:T\to R_j$ be the coordinate ring homomorphism. Set $I_j={\rho }_j(I)$. Each $I_j$ is an ideal: it is an additive subgroup, and multiplying a lift in $I$ on either side by a tuple with prescribed $j$th coordinate proves closure under left and right multiplication by $R_j$.

Certainly $I\subseteq {\prod }_j{I}_j$. Conversely, let $a=(a_1,\dots ,a_n)\in {\prod }_j{I}_j$. For each $j$, choose $b^{(j)}\in I$ with $j$th coordinate $a_j$, and let $e_j$ have $1_{R_j}$ in coordinate $j$ and zero elsewhere. Since $I$ is an ideal, $e_j{b}^{(j)}\in I$, andThereforeThe finite sum is essential to this proof of ideals of a finite product of rings.

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.