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Let $R$ be a UFD and $f_1,\dots,f_n$ be irreducible elements of $R$. Does it follows that the ideal $\langle f_1,\dots,f_n\rangle$ is a prime ideal?. If it's not true in general then is it true in $k[x_1,\dots,x_n]$ where $k$ is an algebraically closed field.

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For future reference, in $\LaTeX$ mode < is for the relation (usually) and \langle is for the angle brackets. –  Asaf Karagila Oct 23 '11 at 17:04
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up vote 8 down vote accepted

No. E.g. suppose that $R = k[x,y],$ with $k$ algebraically closed. Then saying that $f_i$ is irreducible is the same as saying that it cuts out an irreducible affine curve. Then the ideal $\langle f_1, f_2\rangle$ pertains to the intersection of curves $f_1 = 0$ and $f_2 = 0$, which will typically not be irreducible unless $f_1$ and $f_2$ are associate (i.e. scalar multiples of one another, so that they cut out the same curve), or are both of degree $1$, i.e. equations cutting out lines. (I say typically because we are working in the affine plane rather than the projective plane, and so some of the intersections of the intersection points of the two curves might be at infinity, and hence invisible to the ideal $\langle f_1,f_2 \rangle$.)

For a concrete counterexample, choose any non-associate $f_1$ and $f_2$ with at least one of them non-linear, such that none of the intersection points of the corresponding curves lie at infinity, e.g. $f_1 = x^2 - y$ and $f_2 = x-y$. Then $\langle f_1,f_2\rangle = \langle x(x-1), x-y \rangle,$ which is not prime. (The intersection of the parabola $y =x^2$ and the line $y = x$ consists of the two points $(0,0)$ and $(1,1)$.)

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No, for example, in $R=\mathbb{Z}$, 2 and 3 are irreducible, but $$\langle 2,3\rangle=\mathbb{Z}$$ is not prime. A counterexample to the special case you asked about would be $R=k[x]$, with $f_1=x$ and $f_2=x+1$, because $$\langle x,x+1\rangle=k[x]$$ is not prime.

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