Units Of The Ring R = K[x,y,z]/(xyz-xy-xz-yz)

Alright, let's dive into figuring out the units of the ring $R = k[x, y, z]/(xyz - xy - xz - yz)$ where $k$ is a field. This is a cool problem that combines our understanding of polynomial rings and quotient rings, and it requires a bit of algebraic maneuvering. So, let's break it down step by step.

Understanding the Ring R

First, let's get a solid grip on what $R$ actually is. We start with the polynomial ring $k[x, y, z]$, which is just the set of all polynomials in the variables $x, y$, and $z$ with coefficients in the field $k$. Think of $k$ as your usual fields like the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$, or even a finite field. Now, we're taking a quotient of this ring by the ideal generated by the polynomial $xyz - xy - xz - yz$. This means we're essentially saying that in our new ring $R$, the expression $xyz - xy - xz - yz$ is equal to zero.

In other words, in $R$, we have the relation $xyz = xy + xz + yz$. This relation is key to understanding how elements in $R$ behave. Elements of $R$ are polynomials in $x, y, z$, but we can always use this relation to reduce higher-degree terms. For instance, if we see an $xyz$ term, we can replace it with $xy + xz + yz$. This reduction is crucial because it helps us simplify expressions and identify units.

So, to recap, $R$ consists of polynomials in $x, y, z$ with coefficients in $k$, subject to the condition that $xyz = xy + xz + yz$. This condition is what makes the ring $R$ interesting, and it will play a central role in determining its units.

What are Units?

Before we hunt for units in $R$, let's clarify what a unit is. In any ring, a unit is an element that has a multiplicative inverse. That is, an element $u \in R$ is a unit if there exists another element $v \in R$ such that $uv = 1$, where $1$ is the multiplicative identity in the ring. In our case, the multiplicative identity is just the constant polynomial $1$.

In simpler terms, a unit is something you can multiply by another element in the ring to get 1. For example, in the ring of integers $\mathbb{Z}$, the only units are $1$ and $-1$ because they are the only integers that have multiplicative inverses that are also integers. In a field $k$, every nonzero element is a unit because every nonzero element has a multiplicative inverse in the field.

Now, the question is: does $R$ contain any units other than the nonzero elements of $k$? This is what we need to investigate. Elements of $k$ are definitely units in $R$ because $k$ is a field, and every nonzero element in a field has an inverse. But could there be more complicated polynomials in $x, y, z$ that also have inverses in $R$? That's the puzzle we're trying to solve. So, let's proceed with that in mind.

Strategy for Finding Units

Okay, so how do we go about finding these units? Here's a general strategy:

1. Assume that some element $f \in R$ is a unit. This means there exists a $g \in R$ such that $fg = 1$.
2. Consider the possible forms of $f$ and $g$. Since $R$ is a quotient ring of a polynomial ring, $f$ and $g$ can be represented by polynomials in $x, y, z$. We can use the relation $xyz = xy + xz + yz$ to simplify these polynomials.
3. Analyze the equation $fg = 1$. This equation must hold in $R$, which means that $fg - 1$ must be in the ideal generated by $xyz - xy - xz - yz$. In other words, there must exist some polynomial $h \in k[x, y, z]$ such that $fg - 1 = h(xyz - xy - xz - yz)$.
4. Look at the degrees of the polynomials involved. This can often give us valuable information. For example, if $f$ and $g$ are both non-constant polynomials, then $fg$ will have a degree at least 2. But the right-hand side of the equation $fg - 1 = h(xyz - xy - xz - yz)$ must also have degree at least 2, which can give us constraints on $h$.
5. Try some specific examples. Sometimes, the best way to understand a problem is to play around with it. Try plugging in some simple polynomials for $f$ and $g$ and see if you can find a solution to $fg = 1$. This might give you some intuition about what the units in $R$ look like.

Let's try to apply this strategy. Suppose $f$ is a unit in $R$. Then there exists $g \in R$ such that $fg = 1$. We can write $f$ and $g$ as polynomials in $x, y, z$ with coefficients in $k$. If $f$ is just an element of $k$, then it's a unit if and only if it's nonzero. So let's assume $f$ is not just an element of $k$, meaning it involves $x, y,$ or $z$.

Detailed Analysis and Proof

Now, let's dig deeper. Suppose $f \in R$ is a unit, so there exists $g \in R$ such that $fg = 1$. This means that $fg - 1$ is in the ideal generated by $xyz - xy - xz - yz$. Therefore, there exists a polynomial $h(x, y, z) \in k[x, y, z]$ such that

$fg - 1 = h(x, y, z)(xyz - xy - xz - yz)$.

Our goal is to show that if $f$ is a unit, then it must be an element of $k$. Suppose, for the sake of contradiction, that $f$ is not an element of $k$. Then $f$ must involve at least one of the variables $x, y,$ or $z$.

Consider the evaluation map $\phi: R \to k[x, y]$ defined by setting $z = 0$. Note that in $R$, the relation $xyz = xy + xz + yz$ becomes $0 = xy$, which means $xy = 0$. Thus, $\phi$ is well-defined.

Applying $\phi$ to the equation $fg = 1$, we get $\phi(f)\phi(g) = 1$ in $k[x, y]$. But since $k[x, y]$ is an integral domain, this means that both $\phi(f)$ and $\phi(g)$ must be units in $k[x, y]$. The units in $k[x, y]$ are just the nonzero elements of $k$. Thus, $\phi(f)$ and $\phi(g)$ are nonzero constants.

This implies that $f(x, y, 0)$ and $g(x, y, 0)$ are both nonzero constants in $k$. This means that the terms in $f$ and $g$ that don't involve $z$ must be constant. So we can write $f(x, y, z) = c_1 + zf_1(x, y, z)$ and $g(x, y, z) = c_2 + zg_1(x, y, z)$ where $c_1, c_2 \in k$ are nonzero constants, and $f_1, g_1$ are some polynomials in $x, y, z$.

Plugging these into $fg = 1$, we have

$(c_1 + zf_1(x, y, z))(c_2 + zg_1(x, y, z)) = 1$.

Expanding this, we get

$c_1c_2 + c_1zg_1(x, y, z) + c_2zf_1(x, y, z) + z^2f_1(x, y, z)g_1(x, y, z) = 1$.

This means that $c_1c_2 = 1$ and $c_1g_1 + c_2f_1 + zf_1g_1 = 0$ in $R$. From the second equation, we have $c_1g_1 + c_2f_1 = -zf_1g_1$.

Now, suppose $f_1$ is nonzero. Since $c_1, c_2$ are nonzero constants, $f_1$ must involve either $x$ or $y$. If $f_1$ involves $x$ or $y$, then the left-hand side will also involve $x$ or $y$. But the right-hand side is $-zf_1g_1$, which must also involve $x$ or $y$. This implies that $f_1g_1$ must involve $x$ or $y$. However, since $xyz = xy + xz + yz$ in $R$, the presence of $z$ doesn't change the fact that $f_1g_1$ must involve either $x$ or $y$.

To proceed further, it's helpful to use another evaluation map. Consider the map $\psi: R \to k[x]$ defined by setting $y = 0$ and $z = 0$. In $R$, the relation $xyz = xy + xz + yz$ means that if $y = 0$ and $z = 0$, then $0 = 0$, so this map is well-defined.

Applying $\psi$ to $fg = 1$, we get $\psi(f)\psi(g) = 1$. This implies that $f(x, 0, 0)$ and $g(x, 0, 0)$ are both nonzero constants. This tells us that $f$ and $g$ have constant terms, but no terms involving only $x$.

Let's consider the case where $f = f(x, y, z)$ is a unit in $R$. We have $fg = 1$ for some $g \in R$. If we consider the total degree of $f$ and $g$, we run into issues because the relation $xyz = xy + xz + yz$ is not degree-preserving. However, we can still use this relation to simplify the polynomials.

Assume $f$ is not in $k$. Then $f$ must have some terms involving $x, y,$ or $z$. If $f$ contains a term like $x$, then $fg$ would contain a term like $xg$. But $fg = 1$, which has no terms involving $x$. This is a contradiction unless $g$ has a term that cancels out the $x$ term in $f$. However, the only way to cancel out the $x$ term is if $g$ has a term with a negative power of $x$, which is not possible in a polynomial ring.

From our analysis, we can conclude that the only units in $R$ are the nonzero elements of $k$.

Conclusion

Therefore, the ring $R = k[x, y, z]/(xyz - xy - xz - yz)$ contains no units other than the nonzero elements of $k$. We arrived at this conclusion by carefully analyzing the structure of $R$, considering evaluation maps, and applying the relation $xyz = xy + xz + yz$. It's a pretty neat result, showing how the specific form of the quotient relation tightly constrains the units in the ring.