I’m proving the following rings are not isomorphic $$ \mathbb{K}[x,y]\not\cong\mathbb{K}[x,y,z]/(xy-z^2) $$ $\mathbb{K}$ is just an algebraic closed field
I can see they are not isomorphic since there is some polynomials in the latter with $z$, since the higher degrees in $z$ are replaced by $xy$
I think there is a isomorphism:
$$ \mathbb{K}[x,y]\oplus\mathbb{K}[x,y]\cong\mathbb{K}[x,y,z]/(xy-z^2) $$
But how to prove it in formal, thanks in advance for any help!
Question: "But how to prove it in formal, thanks in advance for any help!"
Answer: In fact, much can be said about this question - about the algebra and geometry. Let $K$ be the field of real numbers (or any algebraically closed field) and let $f:=xy-z^2$. We may look at the "algebraic variety" $Z(f)$ defined by $f$ in real 3-space: $V:=\mathbb{R}^3$: $Z(f)$ is the set of points $(a,b,c)\in V$ such that $f(a,b,c)=0$. Geometrically you may view $Z(f)$ as a surface in 3-space, and this surface may be "smooth" or "singular". The singular points of the surface are the points defined by the "Jacobian ideal" $J(f)$ of the the polynomial $f$: The ideal $J(f)$ is the ideal generated by the partial derivatives of $f$:
$$\partial f/\partial x:=y, \partial f/\partial y=x, \partial f/\partial z=-2z,$$
and it follows the Jacobian ideal of $f$ is the ideal $J(f)=(x,y,-2z)\cong (x,y,z)$. Hence the ring $A:=K[x,y,z]/(f)$ has a non-regular point at the origin $0:=(0,0,0)$. The corresponding algebraic variety $Z(f)$ is singular. The ring $K[x,y]$ is regular - the corresponding algebraic variety is real 2-space $\mathbb{R}^2$, and this variety is non-singular. For this reason the rings cannot be isomorphic since regularity/non-singularity is preserved under isomorphism. A similar result holds for $\mathbb{K}$. You must look up a book on algebraic varieties. There is a close connection between the ring $B:=K[x,y,z]/(f)$ and the corresponding variety $Z(f)\subseteq \mathbb{R}^3$.
As mentioned by Lahtonen in the comments: If you write down explicit formulas for two commutative rings and ask "Is there an isomorphism of $k$-algebras between these rings?", this question is usually solved using "invariants". In this case the invariant was "non-singularity/non-regularity". You may find this invariant studied in a book on commutative algebra - maybe Eisenbud/Harris book?
Remark: "I think there is a isomorphism
$$K[x,y]\oplus K[x,y]\cong B."$$
Note: Here is an "elementary proof" along the lines suggested in the comments. If $A:=K[x,y]$ we may consider $A[t]$ and $f(t):=t^2-xy \in A[t]$. $f(t)$ is a monic polynomial in $A[t]$ and your ring $B$ satisfies
$$B \cong A[t]/(f(t)).$$
The ring $B$ has a basis consisting of the elements $1, \overline{t}$ where $\overline{t}$ is the equivalence class of $t$ in the quotient. Hence $B$ is a non-trivial integral extension of $A \cong K[x,y]$. Hence there cannot be an isomorphism $B \cong A$. Note moreover that $f$ is an irreducible polynomial hence $B$ is an integral domain and there cannot be an isomorphism
$$\psi:B \cong K[x,y]\oplus K[x,y]$$
of rings: The right side is not an integral domain. But the map $\psi$ is an isomorphism of $A$-modules: $B$ is a free $A$-module of rank $2$ on the elements $1, \overline{t}$. Hence you are correct when you suspect there is such an "isomophism", but it is an isomorphism of modules and not rings.
For interested readers:
Let $S:=Spec(B)$ and $\mathbb{A}^2_K:=Spec(A)$. The canonical map
$$\phi: S \rightarrow \mathbb{A}^2_K$$
is a finite map - it has finite fibers. If $\mathfrak{m}:=(x-a,y-b)$ with $a,b\in K$ is a maximal ideal in $A$ it follows the "fiber ring" $B\otimes_A \kappa(\mathfrak{m})$ satisfies the following:
$$B\otimes_A \kappa(\mathfrak{m})\cong K[t]/(t-u)\oplus K[t]/(t+u) \cong K \oplus K$$
if $u^2=ab\neq 0$. Here $\kappa(\mathfrak{m}):=A/\mathfrak{m}$ is the residue field. This is beacuse $K$ is algebraically closed and there is an $u\in K$ with $u^2=ab$, hence $t^2-ab=t^2-u^2=(t-u)(t+u)$. The isomorphism is an application of the chinese remainder theorem. If $ab=0$ it follows
$$B\otimes_A \kappa(\mathfrak{m}) \cong K[t]/(t^2):=K[\epsilon]$$
if $ab=0$. If $Z:=V(xy) \subseteq Spec(A)$ and $U:=Spec(A)-Z$, it follows for any closed point $p\in Z$ the fiber $\phi^{-1}(p)$ is one non-reduced point. For a closed point $p\in U$ it follows the fiber $\phi^{-1}(p)$ consists of two points. Generically the morphism $\phi$ has two points in the fiber, but above the closed subscheme $Z$ there is one non-reduced point of "multiplicity two": $dim_K(K[\epsilon])=2$.
Note: If there was an isomorphism $A \cong B$ this would imply that for any maximal ideal $\mathfrak{m}:=(x-a,y-b) \subseteq A$ there was an isomorphism
$$A/\mathfrak{m} \cong B/\mathfrak{m}$$
but
$$A/\mathfrak{m} \cong K\text{ and }B/\mathfrak{m} \cong K[t]/(t^2-ab) \neq K.$$
Hence the calculation above gives an "elementary" proof that $A \neq B$.
