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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Example 10.26.3. In this example we describe $X = \mathop{\mathrm{Spec}}(k[x, y])$ when $k$ is an arbitrary field. Clearly $(0)$ is prime, and any principal ideal generated by an irreducible polynomial will also be a prime since $k[x, y]$ is a unique factorization domain. Now assume $\mathfrak p$ is an element of $X$ that is not principal. Since $k[x, y]$ is a Noetherian UFD, the prime ideal $\mathfrak p$ can be generated by a finite number of irreducible polynomials $(f_1, \ldots , f_ n)$. Now, I claim that if $f, g$ are irreducible polynomials in $k[x, y]$ that are not associates, then $(f, g) \cap k[x] \neq 0$. To do this, it is enough to show that $f$ and $g$ are relatively prime when viewed in $k(x)[y]$. In this case, $k(x)[y]$ is a Euclidean domain, so by applying the Euclidean algorithm and clearing denominators, we obtain $p = af + bg$ for $p, a, b \in k[x]$. Thus, assume this is not the case, that is, that some nonunit $h \in k(x)[y]$ divides both $f$ and $g$. Then, by Gauss's lemma, for some $a, b \in k(x)$ we have $ah | f$ and $bh | g$ for $ah, bh \in k[x]$. By irreducibility, $ah = f$ and $bh = g$ (since $h \notin k(x)$). So, back in $k(x)[y]$, $f, g $ are associates, as $\frac{a}{b} g = f$. Since $k(x)$ is the fraction field of $k[x]$, we can write $g = \frac{r}{s} f $ for elements $r , s \in k[x]$ sharing no common factors. This implies that $sg = rf$ in $k[x, y]$ and so $s$ must divide $f$ since $k[x, y]$ is a UFD. Hence, $s = 1$ or $s = f$. If $s = f$, then $r = g$, implying $f, g \in k[x]$ and thus must be units in $k(x)$ and relatively prime in $k(x)[y]$, contradicting our hypothesis. If $s = 1$, then $g = rf$, another contradiction. Thus, we must have $f, g$ relatively prime in $k(x)[y]$, a Euclidean domain. Thus, we have reduced to the case $\mathfrak p$ contains some irreducible polynomial $p \in k[x] \subset k[x, y]$. By the above, $\mathfrak p$ corresponds to a prime in the ring $k[x, y]/(p) = k(\alpha )[y]$, where $\alpha $ is an element algebraic over $k$ with minimum polynomial $p$. This is a PID, and so any prime ideal corresponds to $(0)$ or an irreducible polynomial in $k(\alpha )[y]$. Thus, $\mathfrak p$ is of the form $(p)$ or $(p, f)$ where $f$ is a polynomial in $k[x, y]$ that is irreducible in the quotient $k[x, y]/(p)$.


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