There's all this basic stuff about $\mathbb{R}[x,y]$ that I don't know, and this impacts negatively on my ability to teach e.g. Year 10 and 11 mathematics effectively (for example).
So let $R$ denote a ring. A well-known result says that if $R$ is a UFD, then $R[x]$ is, too. This tells us that $\mathbb{R}[x,y]$ is a UFD. So prime elements are the same as irreducible elements, and every element of $\mathbb{R}[x,y]$ factors uniquely as a product of such elements. That's about all I know. So, here's a few questions designed to help me patch up my knowledge.
Are the irreducible elements in $\mathbb{R}[x,y]$ well-known and easy to describe, and if so, what are they? When one variable is concerned, the discriminant tells us everything we need to know, of course.
It would seem that both the "circle polynomial" $x^2+y^2-1$ and the "hyperbola polynomial" $x^2-y^2-1$ are both irreducible in $\mathbb{R}[x,y]$. Is that correct? If so, how do we know?
At the risk of asking a meaningless question:) assuming they're both irreducible, why is zero-locus of $x^2+y^2-1$ connected, but the zero-locus of $x^2-y^2-1$ is not?
More generally, can we say anything about the zero-loci of irreducible polynomials that, say, a high school student can understand?
Does homogenization somehow help us? If so, how?
(1) The polynomial ring $K[x]$ over a field $K$ is a Euclidean ring, that is one can perform divisions with remainder. Therefore irreducible and prime elements are one and the same. But observe that even in this case irreducibles are in general not easy to describe: consider the case $K=\mathbb{Q}$. The structure of the field $K$ plays a role. For $K=\mathbb{R}$ things are simple because of the fundamental theorem of algebra and the fact that the complex numbers are an extension of degree 2 of the reals.
The polynomial ring $\mathbb{R}[x,y]$ can be considered as a subring of the polynomial ring $\mathbb{R}(x)[y]$, where $\mathbb{R}(x)$ is the field of rational functions in $x$ with real coefficients, a field that is in a sense as complex as the rational numbers. Irreducibles of $\mathbb{R}(x)[y]$ are therefore not easy to describe.
On the other side a result of Gauß says: a polynomial $f=a_ny^n+a_{n-1}y^{n-1}+\ldots +a_0\in\mathbb{R}[x,y]$, $a_k\in\mathbb{R}[x]$ relatively prime, is irreducible if and only if it is irreducible in $\mathbb{R}(x)[y]$.
(2) A nice check for irreducibility is Eisenstein's criterion: let $A[y]$ be a polynomial ring in $y$ over the UFD $A$. If the coefficients $a_k$ of the polynomial $f=y^n+a_{n-1}y^{n-1}+\ldots +a_0\in A[y]$ are divisible by a prime $p\in A$ and $a_0$ is not divisible by $p^2$, then $f$ is irreducible.
Apply this for $A=\mathbb{R}[x]$ and $p=x-1$ and $p=x^2+1$ to obtain the irreducibility of $y^2+x^2-1$ and $y^2-x^2-1$.
(3) Suppose you draw an irreducible $f$ randomly from $\mathbb{R}[x]$. Then it may or may not have a real root resp. $n$ real roots, $n$ the degree of $f$. Typically some roots are non-real complex numbers. Similar for the zero locus of an irreducible $f\in\mathbb{R}[x,y]$: the zero locus $C(f)$ of $f$ in $\mathbb{C}\times\mathbb{C}$ is always an infinite set - a so-called plane algebraic curve. We have the maps
$ \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{C}\times\mathbb{C},\; (a_1,a_2)\mapsto (a_1+0i,a_2+0i) $
and
$ \mathbb{C}\times\mathbb{C}\rightarrow\mathbb{R}^2\times\mathbb{R}^2,\; (a_1+ib_1,a_2+ib_2)\mapsto (a_1,b_1,a_2,b_2). $
Using them we can consider the zero locus $C(f)$ as a subset of $\mathbb{R}^4$. The intersection
$ C(f)\cap\{(a_1,0,a_2,0) : a_1,a_2\in\mathbb{R}\} $
consists of the points in $C(f)$ having real coordinates ("real zeros"). Depending on the geometric position of $C(f)$ in the space $\mathbb{R}^4$ this intersection may or may not be empty resp. connected.
(4) An interesting fact for high school students: let $f,g\in\mathbb{R}[x,y]$ be two irreducible polynomials such that $f\neq cg$ for all $c\in\mathbb{R}$, then the intersection of their zero loci is finite and bounded from above by $\deg(f)\deg(g)$. Or in other words: the system of polynomials equations
$f(x,y)=0, g(x,y)=0$
has at most $\deg(f)\deg(g)$ solutions.