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I am working on the following problem and am a little stuck on part c). Also if you could verify my other parts, it would be much appreciated.

Edit update: I have figured out part c) and am now struggling with part d).

The goal of this problem is to compute in $\mathbb{A}^2$ the set of zeros of an ideal generated by two irreducible polynomials.

Suppose $k$ is a field and $f$ and $g$ are irreducible polynomials that do not divide each other.

a) Show that if $(f,g)=(1)$, then $\mathcal{Z}((f,g)) = \emptyset$.

My Answer:

Assume the hypothesis. Recall $(1)=\{h\cdot 1|h\in k[x,y]\}=k[x,y]$ and $$ \mathcal{Z}((f,g))=\mathcal{Z}((1))=\{(x,y)\in\mathbb{A}^2\text{ }|\text{ }h(x,y)=0\hspace{3mm}\forall\text{ } h\in (1)\}. $$ Consider $h(x,y)=1\in k[x,y]$. Notice $h(x,y)$ has no zeros. Therefore there does not exist any $(x,y)\in\mathbb{A}^2$ where $h(x,y)=0$ for all $h\in k[x,y]$. Thus $\mathcal{Z}((f,g))=\emptyset$. Done.

b) Assume now $(f,g) \neq (1)$. Consider the ring $R = k[x]$ and $F$ its field of fractions. Apply Gauss lemma (and explain why it is correct to use this lemma) to conclude that $f$ and $g$ are relatively prime in $F[y]$.

My Answer:

Assume the hypothesis. Let $R=k[x]$ and let $F$ be its field of fractions. Because $k$ is a field, we know $R=k[x]$ is a Euclidean Domain. Then $R=k[x]$ is a unique factorization domain (UFD). Therefore we can apply Gauss' lemma. We want to show $f$ and $g$ are relatively prime in $F[y]$.

Notice $R[y]=k[x][y]=k[x,y]$ and $F[y]$ is its field of fractions. Because $f$ and $g$ are irreducible polynomials in $k[x,y]=R[y]$, we know $f$ and $g$ are irreducible in $F[y]$ by Gauss' lemma.

By definition $F[y]$ is a field, then $F[y]$ is a principle ideal domain (PID). By a proposition, we know every nonzero element in a PID is prime $\Leftrightarrow$ it is irreducible. Therefore $f$ and $g$ are prime in $F[y]$. Since $f$ and $g$ do not divide each other, we know $\gcd(f,g)=1$ in $F[y]$.

Thus $f$ and $g$ are relatively prime in $F[y]$. Done.

c) Using b) show that there is a nonzero polynomial in $k[x]$ that is contained in the ideal $(f,g)$.

My Attempt:

Assume the hypothesis. By definition $F[y]$ is a field, then $F[y]$ is a Euclidean Domain. Because $f,g$ are nonzero elements in $F[y]$ and because $f,g$ are relatively prime by part b), we know $(f,g)=(1)$ by a theorem. In particular, there exists $s,t\in F[y]$ such that $$ 1=fs+gt. $$ Note $k[x]\subseteq k[x,y]\subseteq F[y]$. Now we need to find a nonzero polynomial in $k[x]$ that is contained in $(1)=\{1\cdot h\text{ }|\text{ }h\in F[y]\}$.

Here is where I get stuck. I thought about considering the cases when $a,b\in k[x]$ and when $s,t\notin k[x]$. But the problem is I don't know that $f,g\in k[x]$ because of my containment statement above. Any ideas on how to proceed would be much appreciated.

Addition after hint:

Notice $k[x]\subseteq R[y]=k[x][y]=k[x,y]\subseteq F[y]$. Therefore any polynomial in $k[x]$ will be contained in the ideal $(f,g)=F[y]$. Consider $s,t\in F[y]$. Notice \begin{eqnarray} s&=& j_n y^n + j_{n-1}y^{n-1}+\cdots + j_1 y + j_0 \end{eqnarray} and \begin{eqnarray} t&=& h_m y^m + h_{m-1}y^{m-1}+\cdots + h_1 y + h_0 \end{eqnarray} where $j_i\in F$ and $h_l\in F$ for all $i\in\{0,\ldots,n\}$ and for all $l\in\{0,\ldots,m\}$. Then for all $i$ and $l$, \begin{eqnarray} j_i = \frac{a_i}{b_i}\hspace{5mm}\text{and}\hspace{5mm}h_l = \frac{c_l}{d_l} \end{eqnarray} where $a_i,b_i,c_l,d_l\in R=k[x]$ such that $b_i,d_l\geq 0$. Let $b=lcm(b_i)$ and $d=lcm(d_l)$. Notice $b$ and $d$ are nonzero. Now let $e=lcm(b,d)$. Notice $e$ is nonzero. Consider \begin{eqnarray} e(1=fs+gt)\\ e=esf+egt \end{eqnarray} By our choice of $e$, we have $es,eg\in k[x,y]$ because $e$ cancels all the denominators in the $j_n$ and $h_m$ coefficients of $s$ and $t$. Therefore $e$ is a nonzero polynomial of $k[x]$ that is contained in the ideal of $(f,g)$ in $k[\mathbb{A}^2]$. Done.

d) Conclude from the above that in $\mathbb{A}^2$: $$\mathcal{Z}((f,g))=\emptyset \mbox{ or finite.}$$

My Attempt:

Assume hypothesis. Suppose $(f,g)=(1)$ in $\mathbb{A}^2$. Then from part a), we know $\mathcal{Z}((f,g))=\mathcal{Z}((1))=\emptyset$.

Suppose $(f,g)\neq(1)$ in $\mathbb{A}^2$. Then by part c), we know there exists a nonzero polynomial in $k[x]$ that is contained in the ideal $(f,g)$.

Stuck here now... Thanks for taking the time to help me.

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    $X^2+1$ has zeroes, since your field $k$ is probably algebraically closed. – 2017-02-12
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    @MarianoSuárez-Álvarez Thanks for catching that. I assumed $k$ was infinite but that was a bad assumption. I changed $h(x,y)=x^2+1$ to $h(x,y)=1$ in part a). – 2017-02-12

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Since $a,b$ are in $F[y]$, we can find common denominators $c,d$ in $k[x]$ and we can conclude that $ac$ and $bd$ are in $R[y]=k[x,y]$. So multiplying $1=af+bg$ by $cd$ gives us: $cd=(acd)f+(bcd)g$.

Obviously $cd$ is in $k[x]$ and $(acd),(bcd)$ both are in $R[y]$. So we can conclude that $cd$ is indeed in $(f,g)$.

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    Hi. I am still I little confused. Since $a,b\in F[y]$, we know $\frac{e}{c}=a$ and $\frac{f}{d}=b$ where $c,d,e,f\in R[y]$ and $c,d\neq 0$. By how does this give us the common denominator in $k[x]$? – 2017-02-14
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    you can write $a$ as $y^n+j_1y^{n-1}+...+j_n$ with $j_i$ in $F=k(x)$. So you can write every $j_i$ as $j_i=\frac{l_i}{m_i}$ with $l_i,m_i$ in $k[x]$. Now you can choose $c$ as product over all $m_i$ and apply the same idea on $b$ to get the $d$ – 2017-02-14
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    Thank you so much for that hint. It cleared up a lot of questions I had. I was really struggling with how to represent $a$ and $b$. I am stuck now on part d. I believe it has something to do this polynomials of degree $n$ only having finite roots, but I don't know how to appropriately apply this fact. – 2017-02-14
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    for the last part you can assume that the set is not empty, thus you can apply b) and c). This means you can find h(x) in the Ideal $(f,g)$, i.e. if $(f,g)$ vanishes on a point, so must $h(x)$. Thus you get only finite amouny of possibilties for $x$, namely the roots of $h$, with further work I think you can conclude that you also then have only finitely many possibilties left for the $y$-coordinate and you are done. – 2017-02-14
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    Ok so for the $x$-coordinate I could use an argument like this. Suppose $(f,g)\neq(1)$ in $\mathbb{A}^2$. Then by part c), we know there exists a nonzero polynomial, $e$, in $k[x]$ that is contained in the ideal $(f,g)$ of $\mathbb{A}^2$. Additionally from part c), we know that $e=esf+egt$. Therefore if $f(x,y)=g(x,y)=0$, we have $e(x)=0$ in $(f,g)$. Since $e$ is a polynomial of a certain degree, say $r$, we know $e$ will have at most $r$ zeros in $k[x]$. Therefore the $x$-coordinate has a finite number of zeros in $(f,g)$ of $\mathbb{A}^2$. – 2017-02-14
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    For the $y$-coordinate though. I am assuming I will have to look at $F[y]$. Here is what I am thinking so far... Consider $h\in F[y]$. Notice $$ h(y)= j_n y^n + j_{n-1}y^{n-1}+\cdots + j_1 y + j_0 $$ where $j_i\in F$ for all $i\in\{0,\ldots,n\}$. Then for all $i$, we have $j_i = \frac{a_i}{b_i}$ where $a_i,b_i\in R=k[x]$ such that $b_i\geq 0$. Then I need to consider when $h(y)=0$. Right? – 2017-02-14
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    This might work, but I am not sure. I am not seeing a "nice" way right now. You might want to consider $f_i(y)=f(x_i,y)$, $x_i$ being a root of our $e$, which is a polynomial in k[y] and ebenda for $g$. If you can show that $f_i$ and $g_i$ are never equal to 0 at the same time, you are done. – 2017-02-15
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    Can't I just say that since $f$ and $g$ are relatively prime in $F[y]$ so $f_i$ and $g_i$ will never equal 0 at the same time in $F[y]$. Therefore they won't in $k[x,y]$ either? – 2017-02-15
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    Why does they never being 0 at the same time mean that I am done? – 2017-02-15
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    Snce that means that e.g. $g_i(x_i,y)$ only has a finite number of roots in y. And yes if both vanish of $x_i$, both $f_i$ and $g_i$ have $(x-x_i)$ as factor and thus arent coprime anymore – 2017-02-15
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    Gotcha. I think understand the solution to this problem now. Thank you. – 2017-02-15