ProblemSet 1 -- Commutative rings & polynomials

1. Let $p,q\in \mathbb{Z}$ be integers and consider the assignment $$\varphi :\mathbb{Z}\to \mathbb{Z}/⟨p⟩\times \mathbb{Z}/⟨q⟩$$ given for $a\in \mathbb{Z}$ by the rule $\varphi (a)=([a],[a])=(a+⟨p⟩,a+⟨q⟩).$

   1. Show that $\varphi$ is a ring homomorphism. ¹

   2. Assume that $gcd(p,q)=1$. We will later show that there are integers $u,v\in \mathbb{Z}$ for which $up+vq=1.$ Use this to show that $\varphi$ is surjective in this case.

   3. Show that $\varphi$ is not surjective when $p=q$.

2. Consider the set $R=\{a+b\sqrt{5}\mid a,b\in \mathbb{Q}\}$.

   1. Show that $R$ is a ring by showing that $R$ is a subring of $\mathbb{R}$. ²

   Consider the polynomial $f(T)=T^2-5\in \mathbb{Q}[T].$ It is a fact that $$(\mathrm{♣})f(\alpha )={\alpha }^2-5\ne 0\text{for every }\alpha \in \mathbb{Q}\text{.}$$ (We’ll have efficient arguments for this later on).

   2. Use the result $(\mathrm{♣})$ to show for $a,b\in \mathbb{Q}$ that $a+b\sqrt{5}=0⟹a=b=0$.

   3. Use the result $(\mathrm{♣})$ to show that if $0\ne \alpha \in R$ then ${\alpha }^{-1}={\displaystyle \frac{1}{\alpha }}\in R$.

3. A commutative ring $R$ is called a field if every non-zero element of $R$ has a multiplicative inverse.

   A non-zero element $x$ of a commutative ring $R$ is called a zero-divisor if there is a non-zero element $y\in R$ for which $xy=0$.

   1. If $F$ is a field, show that $F$ contains no zero divisors.

   2. A commutative ring with no zero divisors is called an integral domain. Show that any subring of a field is an integral domain. Conclude that $\mathbb{Z}$, $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{5}]$ are all integral domains.

   3. If $R$ is any commutative ring (with identity), show that the cartesian product $R\times R$ is a commutative ring which is never an integral domain.

   4. Show for any integer $n>1$ that the ring $\mathbb{Z}/⟨n^2⟩$ is not an integral domain.

4. Let $F$ be a field and let $F[T]$ be the polynomial ring in the variable $T$ with coefficients in $F$.

   Then $F[T]$ is a vector space over $F$ (in the sense of linear algebra), and the set of monomials $\{1,T,T^2,\cdots ,T^n,\cdots \}$ is a basis for this vector space.

   Let $\varphi :F\to F$ be a ring isomorphism ³, and define a mapping $\mathrm{\Phi }:F[T]\to F[T]$ by the rule $$\mathrm{\Phi }\left(\sum _{i=0}^N{a}_i{T}^i\right)=\sum _{i=0}^N\varphi (a_i)T^i.$$

   1. Show that $\mathrm{\Phi }$ is a ring homomorphism.

   2. Show that $\ker (\mathrm{\Phi })=\{0\}$ ⁴ and conclude that $\mathrm{\Phi }$ is injective.

   3. Show that $\mathrm{\Phi }$ is surjective as well. Thus $\mathrm{\Phi }$ is an isomorphism of rings (i.e. $\mathrm{\Phi }$ is an automorphism of the ring $F[T]$).

      (Hint You can argue the surjectivity directly. Or you can argue that the image of $\mathrm{\Phi }$ is a vector subspace of $F[T]$ containing the basis $\{T^i\}$).