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Are the ideals in $\mathbb{C}[X,Y]$, generated by the polynomials $X+Y$ and $XY$ prime or maximal ? These (among others) were recently mentioned as examples of prime and maximal ideals in class, but I didn't keep up and now I can't figure out which is which.

My hunch would be that $XY$ isn't prime, since I can't "separate" for example the $X$ by doing additive operations in $\left$ and multiplying $XY$ with elements from $\mathbb{C}[X,Y]$, but I couldn't find a specific polynomial that works. But for maximality, as well as for the other ideal I have no idea(l) ;)

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    Note that $(XY) = (X)(Y)$; when is a principal ideal of a polynomial ring prime? This should help you with $(X+Y)$.2012-10-16

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$XY\in(XY)$, but neither $X\in(XY)$ nor $Y\in (XY)$, hence $(XY)$ is not prime (and even less maximal).

$(X+Y)$ is the kernel of the homomorphism $\phi\colon\mathbb C[X,Y]\to \mathbb C[X]$ induced by $X\mapsto X$, $Y\mapsto -X$ and $\mathbb C[X]$ has no zero-divisors (but is not a field). Hence $(X+Y)$ is prime (but not maximal).


To see that indeed $\ker\phi=(X+Y)$ (with $\supseteq$ being trivial), consider a nonzero element $f(X,Y)\in\ker\phi$ of minimal degree $n$ in $Y$. Collect all monomials of same degree in $Y$ and thus write $f(X,Y) = g_0(X)+Yg_1(X)+ \ldots + Y^{n-1}g_{n-1}(X)+ Y^ng_n(X).$ If $n=0$, then $\phi(f)=g_0=0$ implies $f=0$. Therefore $n\ge1$. If we add a multiple of $X+Y$ to $f$, we obtain another element of $\ker\phi$, for example $f(X,Y)-(X+Y)Y^{n-1}g_n(X) = g_0(X)+Yg_1(X)+ \ldots + Y^{n-1}(g_{n-1}(X)-Xg_n(X)).$ As this has at most degree $n-1$ in $Y$ it must be the zero polynomial by minimality of $n$, that is $g_0=\cdots=g_{n-2}=0$ and $g_{n-1}=Xg_n$. But then $f(X,Y)=Y^{n-1}(X+Y)g_n(X)$ is a multiple of $X+Y$.

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    @user16008: that is$a$general property of polynomial rings. The kernel of the "evaluation at $a$" homomorphism $R[t] \to R$ is precisely the ideal generated by $(t-a)$, which corresponds to the *congruence* relation generated by $t \equiv a$. Depending on how you think of things, factoring $t^n - a^n$ might help.2012-10-24