ProblemSet 1 -- Commutative rings & polynomials

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

    1. Show that \(\phi\) 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 \(\phi\) is surjective in this case.

    3. Show that \(\phi\) 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 \[(\clubsuit) \quad f(\alpha) = \alpha^2 - 5 \ne 0 \quad \text{for every $\alpha \in \mathbb{Q}$.}\] (We’ll have efficient arguments for this later on).

    1. Use the result \((\clubsuit)\) to show for \(a, b \in \mathbb{Q}\) that \(a + b \sqrt{5} = 0 \implies a = b = 0\).

    2. Use the result \((\clubsuit)\) to show that if \(0 \ne \alpha \in R\) then \(\alpha^{-1} = \dfrac{1}{\alpha} \in R\).

  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 \(\mathbb{Z}/\langle n^2 \rangle\) 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 \(\phi:F \to F\) be a ring isomorphism 3, and define a mapping \(\Phi:F[T] \to F[T]\) by the rule \[\Phi\left(\sum_{i=0}^N a_i T^i\right) = \sum_{i=0}^N \phi(a_i) T^i.\]

    1. Show that \(\Phi\) is a ring homomorphism.

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

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

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

  1. If \(R\) and \(S\) are rings, a function \(\phi:R \to S\) is a ring homomorphism if \(\phi\) is a homomorphism of additive groups and if \(\phi(ab) = \phi(a)\phi(b)\) for every \(a,b \in R\).↩︎

  2. If \(T\) is a ring, a subset \(S\) of \(T\) is a subring provided that \(S\) is an additive subgroup of \(T\) and that \(S\) is closed under the multiplication obtained from \(T\). Notice that if \(S\) is a subring, then \(S\) is itself a ring (under the operations of \(T\)).↩︎

  3. A ring homomorphism is called an isomorphism if it is invertible (as a function). The inverse function is always a ring homomorphism as well.↩︎

  4. Here \(\ker\) just means the kernel of \(\Phi\) viewed as a homomorphism of additive groups.↩︎