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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