Questions about $K[x_1, \ldots , x_4]/(x_1x_2-x_3x_4)$ being a graded ring So I was trying to understand this answer to the question of why $K[x_1, \ldots , x_4]/(x_1x_2-x_3x_4)$ ($K$ being a field) is not an UFD, and the author seems to use the fact that $K[x_1, \ldots , x_4]/(x_1x_2-x_3x_4)$ is a graded ring, where the grading is (I quote) “inherited” from $K[x_1, \ldots , x_4]$.
I’ve never encountered this concept before but I’ve read online that a polynomial ring $R = A[x_1, x_2, \ldots, x_n]$ over a ring $A$ is a graded ring $R=\oplus_{n=1}^{\infty}$ where $R_n$ is the set of homogeneous polynomials of degree $n$.
What I don’t understand is how that structure is inherited to the quotient ring in this context, and how the author of the aforementioned answer seems to use this fact to conclude that if $x_1$ has a factorization into two non-zero non-unit elements in this quotient ring, these must be a polynomial of degree one $\sum_{i=1}^4 a_ix_i$ ($a_i \in K$) and a constant $a\in K$.
 A: The inheritance of the grading comes from looking at the quotient on each $R_d$, because the quotient with a homogenous ideal $I\subset R$ induces a quotient map on each $R_d$ so $R/I = \bigoplus_{d\geq 0}R_d/I_d$, where $I_d$ is the set of all elements in $I$ with degree $d$. So when applying the quotient map you can apply it separately to all elements of a certain degree and then their image is either of the same grade or zero.
If you now take $R=K[X_1,X_2,X_3,X_4]$ and $I = (X_1X_2-X_3X_4)$ this means that if you take the image of a homogenous polynomial (all summands of same degree) it is either again homogenous of the same degree or zero, if it is in $I$. For example conside $X_1^3+X_1^2X_2-X_1X_3X_4 + I = X_1^3+X_1(X_1X_2-X_3X_4)+I = X_1^3+I$
If you now take $x_i = X_i+I \in R/I$ this has again degree one, as it is not contained in $I$. Now for calculations with the degree apply the same rules as usual, because you again have a graded ring, you just have an additional relation on the elements that $x_1x_2 - x_3x_4 = 0$, this however has not really anything to do with the degree of polynomials, as for example in $R$ $X_1X_2-X_1X_2=0$ also zero.
Now as the same rules for degree apply as always you now that if $f*g = x_1$ $1 = deg(x_1) = deg(f*g)=deg(f)+deg(g)$ and as $deg(f),deg(g)\geq 0$ one of them has degree one and one has degree zero.
