show the coordinate axis form an algebraic set in $\Bbb{R}^3$ So by coordinate axis in $\Bbb{R}^3$ im assuming they mean the subspace of $\Bbb{R}^3$:
$$A = \{(x,0,0),(0,y,0),(0,0,z): x,y,z \in \Bbb{R}\}.$$
Or should I write this as
$$\{(x,0,0):x \in \Bbb{R}\}\cup \{(0,y,0):y \in \Bbb{R}\} \cup \{(0,0,z):z \in \Bbb{R}\}$$
Ive noticed that the $x-$ axis is the zero set $V(y,z)$ and the $y-$axis is the zero set $V(x,z)$ and similarly the $z-$axis is the zero set $V(x,y)$ am I headed in the right direction? and does being inside of $V(x,y$) for instance imply $x,y=0$? Also for the zero set associated to (if thats the right terminology) $A$, do I take the union of these $V$'s that I found or? So if anyone wants to help out, is it just $V(y,z) \cup V(x,z) \cup V(x,y)$?
 A: It depends a bit on what an algebraic set means for you. However, the idea is that the union of the three axes is a union of $3$ "irreducible" algebraic sets. A peculiarity of real numbers (as compared with the complex numbers) is that the $x$-axis can be written as $V(y^2+z^2)$: indeed, $(x_0,y_0,z_0)$ satisfies $y^2+z^2=0$ if and only if $y_0=z_0=0$. An analogous argument gives the $y$-axis as $V(x^2+z^2)$ and the $z$-axis as $V(x^2+y^2)$.
If you have a set defined by the equation: $V(P)$ and another defined by $V(Q)$, note that $V(P) \cup V(Q) = \{(x,y,z): P(x,y,z) = 0\:\text{or}\:Q(x,y,z) = 0\}= \{(x,y,z):P(x,y,z)Q(x,y,z)=0\}$.
This says that taking the product of polynomials on the algebraic side corresponds to taking the union of the zero loci geometrically. This tells you that the condition you are after is (as suggested in the comments) given by taking the product of the defining equations for each axis: $V((x^2+y^2)(y^2+z^2)(x^2+z^2))$.
A: A solution which works over $\mathbb{R}$, or any other field too, is the following
\begin{align*}
    V&=V(x,y)\cup V(x,z)\cup V(y,z)\\
    &=V(x^2y,xy^2,y^2z,x^2z,xz^2,yz^2,xyz)
\end{align*}
Where I use $V(I)\cup V(J)=V(IJ)$ where the product $IJ$ of ideals is the set of polynomials $\sum_if_ig_i$ for $f_i\in I$ and $g_i\in J$. If your ideals are $I=(f_1,...,f_n)$ and $J=(g_1,...,g_m)$ then $IJ=(f_1g_1,f_1g_2,...,f_ng_m)$.
