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Let $K$ be a field. Consider the polynomial rings $K[x] = K[x_1, \dots, x_n]$ and $K[y] = K[y_1, \dots, y_m]$. Let $\mathfrak{p} \subset K[x]$ and $\mathfrak{q} \subset K[y]$ be prime ideals. The extensions of these ideals to $K[x, y]$, namely $\mathfrak{p}K[x, y]$ and $\mathfrak{q}K[x, y]$, are also prime ideals because $$ \frac{K[x, y]}{\mathfrak{p}K[x, y]} \cong \left( \frac{K[x]}{\mathfrak{p}} \right)[y]. $$

My question is whether the sum $\mathfrak{p} + \mathfrak{q} \subseteq K[x, y]$ is a prime ideal. I believe this is true, but I am not sure how to prove it.

I know that $$ \frac{K[x, y]}{(\mathfrak{p} + \mathfrak{q})} \cong \frac{\left( \frac{K[x]}{\mathfrak{p}} \right)[y]}{\mathfrak{q} \left( \frac{K[x]}{\mathfrak{p}} \right)[y]}, $$ but since I do not know if $\mathfrak{q}$ is prime in $\left( \frac{K[x]}{\mathfrak{p}} \right)[y]$, I cannot conclude that the quotient is an integral domain.

Heraklit
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1 Answers1

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If $K$ is not algebraically closed, $(\mathfrak p+\mathfrak q)K[x,y]$ need not necessarily be a prime ideal. To see this, let $n=1,m=1$ and $\mathfrak p=(f(x)),\mathfrak q=(g(y))$ with irreducible polynomials $f,g$. Then $K[x]/\mathfrak p=L$ is a finite extension of $K$ and $K[x,y]/(f,g)\cong L[y]/gL[y]$ is an integral domain if and only if $g$ is irreducible over $L$. But this is not necessarily the case, take e.g. $f=g$ of degree $>1$ (which exists if $K$ is not algebraically closed).

If however $K$ is algebraically closed, the claim is true, more generally see here or here

leoli1
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