28. Prove that an ideal $I$ of a ring $R$ is maximal if and only if the quotient ring $R/I$ is a field.

Assume $R/I$ is a field. Let $J$ be an ideal such that $J \supsetneq I$. Since $J \supseteq I, \exists x \in J$ such that $x \notin I (I + n \subset J/I)$. So $I + x \neq I + 0$. Since $R/I$ is a field, $\exists (I + y) \in R/I$ such that:

From the above, we get $xy \sim 1, xy-1 \in I \in J$.

$1 = xy - (xy - 1)$. Thus $\forall n \in R, n = n \cdot 1$, so $J = R$, meaning $I$ is maximized.

29. a. Consider the polynomial ring $\mathbb{Q}[x,y]$ in two variables with rational coefficients. (One way to think about it is as the polynomial ring in variable $y$ whose coefficients come from the single-variable polynomial ring $\mathbb{Q}[x]$.) Find an ideal $I\subseteq \mathbb{Q}[x,y]$ that is prime but not maximal.

Proof: Let $R = \mathbb{Q}[x,y]$ and let $I = (x + 1)$ be the ideal generated by the polynomial $(x + 1)$. $I$ is a rpime ideal of $R$ but not a maximal ideal.

Suppose $f, g \in \mathbb{Q}[x, y]$ and suppose that $fg \in I$. Then $(x + 1)|fg$, say $fg=(x+1)h$. We can write each of $f, g, h$ as a product of irreducible polynomials. Since $\mathbb{Q}[x,y]$ is a unique factorization and since the irreducible polynomial $x + 1$ occurs on one side of the equation, an associate of it must occur on the other side, hence that associate, hence $x + 1$, is a factor of either $f$ or $g$, so $f \vee g$ is in I.