How can I prove that the ideal $ (xy-1) \subset k[x,y] $ is radical? I think that it's enough to prove that the polynomial $xy-1$ is irreducible. How can we prove that?

Here $k$ is a field, I'm not sure if I have to assume that $k$ is algebraically closed, but let's assume it if it's necessary.

  • $\begingroup$ Go by contradiction write $(xy-1)=f(x,y)g(x,y)$ then write $f$ and $g$ as elements of $k[x][y]$ and look at degrees. $\endgroup$ – JSchlather Jan 18 '13 at 1:48
  • 3
    $\begingroup$ Your polynomial $xy - 1$ is a linear polynomial in the ring $k[y]$. $\endgroup$ – user38268 Jan 18 '13 at 1:59
  • $\begingroup$ Your question seems ambiguous: do you want to know why $xy-1$ is irreducible or whether irreducibility implies that the associated prime ideal is radical? Anyway the first interpretation is addressed in the comments and I have given an answer for the other interpretation. $\endgroup$ – Georges Elencwajg Jan 18 '13 at 4:38
  • $\begingroup$ @BenjaLim: Nitpick: the ring $k[x][y]$. (or $k(x)[y]$) $\endgroup$ – user14972 Jan 18 '13 at 7:04

For variety....

An ideal is radical if and only if its quotient ring is reduced: i.e. no nilpotents.

I claim $k[x,y] / (xy-1) \cong k[x, \frac{1}{x}]$. That is, the ring of fractions of $k[x]$ formed by inverting $x$. As this is a subring of $k(x)$, it is reduced.

Why would I claim this? Because of the calculation

$$ xy-1 \equiv 0 \pmod{xy-1} $$ $$xy \equiv 1 \pmod{xy-1} $$ $$ y \equiv \frac{1}{x} \pmod{xy-1} $$

So clearly there is a map $f : k[x,y] \to k[x,\frac{1}{x}]$ defined by $f(x)=x$ and $f(y) = \frac{1}{x}$. Because $f(xy-1) = 0$, it gives a map $k[x,y] / (xy-1)$.

The reverse direction is the map $g : k[x,\frac{1}{x}] \to k[x,y] / (xy-1)$ given by

$$g\left( \frac{a(x)}{x^n} \right) = y^n a(x)$$

First, let's check this is well-defined:

$$g\left( \frac{x^m a(x)}{x^{m+n}} \right) = y^{m+n} x^m a(x) \equiv y^n a(x) = y^n a(x) = g\left( \frac{a(x)}{x^n} \right)\pmod{xy-1} $$

It respects addition:

$$g\left( \frac{a(x)}{x^n} + \frac{b(x)}{x^n} \right) = y^n (a(x)+b(x)) = y^n a(x) + y^n b(x) = g\left( \frac{a(x)}{x^n} \right) + g \left( \frac{b(x)}{x^n} \right)$$

and multiplication:

$$g\left( \frac{a(x)}{x^m} \cdot \frac{b(x)}{x^n} \right) = y^{m+n} (a(x) b(x)) = \left( y^m a(x) \right) \left( y^n b(x) \right)= g\left( \frac{a(x)}{x^m} \right) \cdot g \left( \frac{b(x)}{x^n} \right)$$

and so $g$ is a homomorphism. Then,

$$ f\left(g\left( \frac{a(x)}{x^n} \right)\right) = f\left( y^n a(x) \right) = \frac{a(x)}{x^n} $$

and therefore $g$ is an injective homomorphism. It is surjective, since $g(x) = x$ and $g(1/x) = y$, therefore $g$ is bijective, and thus an isomorphism.


If $A$ is a factorial ring (=UFD), then for $f\in A$ the principal ideal $I=fA $ is radical if and only if, in the decomposition $f=p_1^{a_1}\cdots p_r^{a_r}$ into non-associated irreducibles $p_i$, all multiplicities equal one: $a_i=1$.
This follows from the easy to prove general formula for the nilpotent radical: $$\sqrt I=\sqrt {p_1^{a_1}\cdots p_r^{a_r}A}=p_1\cdots p_rA$$

In particular an irreducible element $p\in A$ generates a radical ideal $pA$ and this applies to your case $p=xy-1\in A=k[x,y]$, since if $k$ is a field (algebraically closed or not), it was proved by Gauss that the polynomial ring in any number of indeterminates $k[x_1\cdots,x_n]$ is a factorial ring.


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