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Say $k$ is a field. Consider the standard graded $k$-algebra $R = k[A,B]$, where $A$ and $B$ are indeterminates. Suppose, for some $x \in R$, that $x^2$ is homogeneous of degree $2$. Does it follow that $x$ is homogeneous of degree $1$?

My attempt to prove this is circular, and asks to use what I am trying to prove. I have bolded that statement which requires what I am trying to prove.

For the sake of contradiction, write $x = p + h$, where $p, h \in R$ and $h$ is a homogeneous element of $R$ with $\text{deg}_R(h) \neq 1$. From this we find $x^2 = p^2 + 2ph + h^2$. By the given information stated above, we see that $\text{deg}_R(x^2) = 2$. Since $h^2$ is a homogeneous summand of $x^2$, it must be the case that $\text{deg}_R(h^2) = 2$. But from this we deduce that $\text{deg}_R(h) = 1$, which is a contradiction.

Hence all homogeneous summands of $x$ must be degree 1, which implies $\text{deg}_R(x) = 1$, as desired.

beanshadow
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1 Answers1

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Yes, because if $x$ and $y$ are homogeneous elements in a graded algebra and $xy \ne 0$ then

$$\deg(xy) = \deg(x) + \deg(y)$$

So assuming that $x^2 \ne 0$,

$$\deg(x^2) = 2\deg(x)$$

so

$$\deg(x^2) = 2 \implies \deg(x) = 1$$

John Baez
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