I'm studying a definition of a resultant so I have to compute the product of two vector spaces $\mathcal{P}^{i}$ with elements that are polynomials of degree less than $i$. If $1+x \in \mathcal{P}^{2}$ and $2+x+x^{2} \in \mathcal{P}^{3}$ then is their element in the product space $\mathcal{P}^{2}\times \mathcal{P}^{3}$ the element $(1,x,2,x,x^{2})$? That is what I guess to do from this slide of a powerpoint
I have to find out how $\mathcal{P}^{2}\times \mathcal{P}^{3} \simeq \mathcal{P}^{5}$..
To write the element in $\mathcal{P}^2\times\mathcal{P}^3$, one would typically write $$ (1+x,2+x+x^2) $$ to indicate how the elements are grouped.
The isomorphism to $\mathcal{P}^5$ isn't natural - it involves a choice. If a chosen basis for $\mathcal{P}^2$ is $\{1,x\}$ and a chosen basis for $\mathcal{P}^3$ is $\{1,x,x^2\}$, then a basis for $\mathcal{P}^2\times\mathcal{P}^3$ is $$ \{(1,0)\;,\; (x,0)\;,\; (0,1)\;,\; (0,x)\;,\; (0,x^2)\}. $$
On the other hand, a basis for $\mathcal{P}^5$ is $\{1,x,x^2,x^3,x^4\}$. Then, one can pair up the basis elements of $\mathcal{P}^2\times\mathcal{P}^3$ with those of $\mathcal{P}^5$.
Therefore, $(1+x,2+x+x^2)$ in $\mathcal{P}^2\times\mathcal{P}^3$ would correspond to $1+x+2x^2+x^3+x^4$, but this isomorphism is only as vector spaces and has limited ring-theoretic properties.