Exercise 1.3

Proof of $i),\mathit{ii})$. We induce on $r$. $r=1$ is Exercise 2. Now consider $A[x_1,\dots ,x_r]$ as a polynomial ring in an indeterminate $x_r$ over the ring $A[x_1,\dots ,x_{r-1}]$, and write $f=\sum <!--\; FUNCTION\; APPLICATION\; -->f_i{x}_r^i$, where $f_i\in A[x_1,\dots ,x_{r-1}]$. By Exercise ?? , $f$ is a unit (resp. nilpotent) if and only if $f_0$ is a unit and $f_1,\dots ,f_n$ are nilpotent (resp. each coefficient $f_i$ is nilpotent). By inductive hypothesis, this is equivalent to the constant term $a_0$ of $f_0$ being a unit and each other coefficient of $f_i$ in $A$ being nilpotent (resp. every coefficient of the $f_i$ in $A$ being nilpotent). □

Proof of $\mathit{iii})$. $\Rightarrow$. Trivial since $a\in A\subset A[x_1,\dots ,x_r]$.

$\Leftarrow$. Suppose not. Using multi-index notation $x^{\alpha }=x_1^{{\alpha }^1}\cdots x_r^{{\alpha }^r}$ where $\alpha =({\alpha }^1,\dots ,{\alpha }^r)\in {\mathbb{N}}^r$, consider the well-ordering $\succ$ on monomials in $A[x_1,\dots ,x_r]$ defined by $\alpha \succ \beta$ if the left-most nonzero entry in $\alpha -\beta$ is positive (the lexicographic order). Define $\mathrm{lt}<!--\; FUNCTION\; APPLICATION\; -->(f)$ to be the leading term of $f$ with respect to $\succ$, and $\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(f)$ to be the coefficient of $\mathrm{lt}<!--\; FUNCTION\; APPLICATION\; -->(f)$. Now write $f=a_n{x}^{{\alpha }_n}+a_{n-1}{x}^{{\alpha }_{n-1}}+\cdots +a_0{x}^{{\alpha }_0}$ where ${\alpha }_n\succ {\alpha }_{n-1}\succ \cdots \succ {\alpha }_0$. Let $g$ be a polynomial $\mathit{fg}=0$ such that $\mathrm{lt}<!--\; FUNCTION\; APPLICATION\; -->(g)$ is minimal with respect to $\succ$. We claim $a_{n-s}g=0$ for all $0\le s\le n$. First, $\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{fg})=\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(f)\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(g)=0$, hence $a_{n}g=0$ since $g\succ a_{n}g$ but $f\cdot a_{n-s_0}g=0$. Now suppose that $a_{n-s}g=0$ for all $0\le s<s_0$. Then, we have that

by inductive hypothesis. Then, $a_{n-s_0}\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(g)=0$, hence $a_{n-s_0}g=0$ since $g\succ a_{n-s_0}g$ but $f\cdot a_{n-s_0}g=0$. By induction, $a_{n-s}g=0$ for all $0\le s\le n$, hence letting $a=\mathrm{lc}<!--\; FUNCTION\; APPLICATION\; -->(g)$ gives $\mathit{af}=0$, a contradiction. □

Proof of $\mathit{iv})$. $\Rightarrow$. Suppose $\mathit{fg}$ is primitive but $f$ is not. Then, the coefficients of $f$ generate an ideal $𝔞\subset 𝔪⊊A$ by Cor.  $1.4$, where $𝔪$ is a maximal ideal. Now consider the natural homomorphism $\phi :A[x_1,\dots ,x_r]\to A∕𝔪[x_1,\dots ,x_r]$ induced by the map $A\to A∕𝔪$ on coefficients. $\phi (f)=0$ implies that $\phi (\mathit{fg})=0$, and so the coefficients of $\mathit{fg}$ are in $𝔪$, a contradiction.

$\Leftarrow$. Suppose $f,g$ are primitive but $\mathit{fg}$ is not. Then, the coefficients of $\mathit{fg}$ generate an ideal $𝔞$, and $𝔞\subset 𝔪⊊A$ by Cor.  $1.4$, where $𝔪$ is again a maximal ideal. The same map $\phi$ defined above has $\phi (\mathit{fg})=0$, and so $\phi (f)\phi (g)=0$, i.e., either $\phi (f)=0$ or $\phi (g)=0$, for if either were a zero divisor, then there exists $a\in A∕𝔪$ such that $\mathit{a\phi }(f)=0$ by $\mathit{iii})$, contradicting that $A∕𝔪$ is a field. Thus, the coefficients of either $f$ or $g$ are contained in $𝔪$, a contradiction. □