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The biggest thing that it stopping me from progressing in this question is the fact that I have two elements in the ideal, this topic it still very new to me.

I can show that $I=\langle x \rangle$ and $I=\langle y \rangle$ are prime ideals on their own by showing that $^{K[x,y]}/_I$ is an integral domain through the First Isomorphism Theorem for Rings. How do I proceed with this question now that there are two elements in the ideal?

If I were to attempt it, knowing that $K$ is a field, I would define a homomorphism $\phi:K[x,y]\rightarrow K,f(x,y)\rightarrow f(0,0)$. Clearly $\ker \phi = \langle x \rangle \cup\langle y \rangle=\langle x,y\rangle$. So we have that $^{K[x,y]}/_{\langle x , y \rangle} \cong K$ by the First Isomorphism Theorem for Rings, and since $K$ is a field, it is an integral domain, and thus $\langle x , y \rangle$ is a prime ideal.

Maksymilian5275
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  • Yes, $(x,y)$ is a maximal ideal, see also here and hence a prime ideal. – Dietrich Burde Apr 20 '21 at 15:33
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    Your formula $\langle x \rangle \cup\langle y \rangle=\langle x,y\rangle$ is false. For example $x+y\in \langle x,y\rangle$ but $x+y\notin \langle x \rangle \cup\langle y \rangle$ – Georges Elencwajg Apr 20 '21 at 15:34
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    Not sure what is your question. Your proof is correct. Though I'd like to point that the equality $(x,y)=(x)\cup (y)$ is false. – Mark Apr 20 '21 at 15:34
  • That's an understanding error on my part, I suppose I was trying to say that the kernel is $\langle x \rangle$ AND $\langle y \rangle$ – Maksymilian5275 Apr 20 '21 at 15:54

2 Answers2

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You can also try to factor out by the ideal $(x,y)$. $$\frac{K[x,y]}{(x,y)} \cong \frac{\frac{K[x,y]}{(x)}}{\frac{(x,y)}{(x)}} \cong \frac{K[y]}{(y)} \cong K$$

cos_dm_math21
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By definiton the ideal $(x,y)$ is the set of polynomials in $K[x,y]$ of the form $$ xA(x,y)+yB(x,y) $$ where $A(x,y)$ and $B(x,y)$ are any polynomials. Thus, it is quite clear that a polynomial $$ a_{0,0}+a_{1,0}x+a_{0,1}y+a_{2,0}x^2+a_{1,1}xy+a_{0,2}y^2+\cdots $$ is in $(x,y)$ if and only if $a_{0,0}=0$.

Thus one may check directly that $(x,y)$ is a prime ideal by showing that the constant term of a product $P(x,y)Q(x,y)$ is $0$ of and only if one of the constant terms of $P(x,y)$ and $Q(x,y)$ is zero. This is true because the constant term of a prduct of polynomials is the prduct of the constant terms and $K$ is a field.

(So the result is true also for the ring $R[x,y]$ where $R$ is a domain, although in this case $(x,y)$ is not maximal)

Andrea Mori
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