Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 3

The first isomorphism theorem for rings states that the mapis a ring isomorphism. It is well defined because two representatives differ by an element of the kernel exactly when their images agree. It preserves addition and multiplication, is onto its stated codomain, and has trivial kernel. In particular, if $f$ is surjective then $R/\ker f\cong S$.

Define $\Phi :R\to R/I\times R/J$ by $\Phi (r)=(r+I,r+J)$. It is a ring homomorphism withSince $I+J=R$, choose $i\in I,j\in J$ such that $i+j=1$. For arbitrary $a,b\in R$, the element $r=aj+bi$ satisfies $r-a=i(b-a)\in I$ and $r-b=j(a-b)\in J$. Hence $\Phi (r)=(a+I,b+J)$, proving surjectivity. The first isomorphism theorem now givesThis is the two-ideal Chinese remainder theorem for ideals. The hypothesis is comaximality of the two ideals; no assertion that every pair of ideals has this property is needed.

Coprimality gives polynomials $a,b\in \mathbb{Q}[x]$ with $ag+bh=1$, so the ideals $(g)$ and $(h)$ are comaximal. AlsoIndeed, if $u=gr=hs$, multiplying $ag+bh=1$ by $u$ writes $u=aghs+bhgr$, a multiple of $gh$; the reverse inclusion is immediate. Part (b), with $f=gh$, givesThe explicit isomorphism takes the class of a polynomial $r$ to its classes modulo $g$ and $h$.

Because $g,h$ are nonconstant, their classes in $\mathbb{Q}[x]/(gh)$ are nonzero: each has smaller degree than $gh$, so neither is a multiple of $gh$. Their product is zero. Thus the quotient has zero divisors.

If both factors have degree one, each quotient by one factor is isomorphic to $\mathbb{Q}$, by evaluation at its rational root. Coprimality ensures the two roots are distinct. Part (c) therefore identifies the ring with $\mathbb{Q}\times \mathbb{Q}$. If $(r,s{)}^2=(0,0)$, then $r^2=s^2=0$ in a field, so $r=s=0$.A ring may therefore have zero divisors without having nonzero nilpotent elements.

Factorization over $\mathbb{Q}$ givesFor $R_1$, the factors $x-1$ and $x+1$ are coprime. The nonzero elements $(1,0)$ and $(0,1)$ multiply to zero, so the ring is neither an integral domain nor a field. Squaring coordinatewise shows it has no nonzero square-zero element.

For $R_2$, the polynomial $x^2+1$ has no rational root and is irreducible of degree two over $\mathbb{Q}$. The quotient is consequently a field, hence an integral domain, and has no nonzero square-zero element.

For $R_3$, the class $\epsilon$ is nonzero by its degree, but ${\epsilon }^2=0$. Thus this ring is neither an integral domain nor a field, and does have a nonzero square-zero element.No two are isomorphic. Being a field separates $R_2$ from the other two; existence of a nonzero square-zero element separates $R_1$ from $R_3$. Both properties are preserved by ring isomorphisms.