Basis of a submodule of $\mathbb{Q}[X]^3$ over $\mathbb{Q}[X]$ I want to find the basis of a submodule of $\mathbb{Q}[X]^3$ generated by $(2X-1, X, X^2+3)$, $(X,X,X^2)$ and $(X+1,2X,2X^2-3)$. I determine that
$$
\begin{pmatrix} 2X-1 & X & X^2+3 \\ X & X & X^2 \\ X+1 & 2X & 2X^2-3\end{pmatrix}
\sim 
\begin{pmatrix} X & X & X^2 \\ 1 & X & X^2-3 \\ 0 & 0 & 0\end{pmatrix}
$$
by basic row operations. Is it valid to multiply the second row by $X$ and subtract the first row from it to yield
$$
\begin{pmatrix} X & X & X^2 \\ 0 & X^2-X & X^3-X^2-3X \\ 0 & 0 & 0\end{pmatrix}?
$$
I'm just a little sure what to do when the entries are polynomials, and whether I can scale a row by $X$, or only units in $\mathbb{Q}[X]$. 
 A: No, you can't multiply by $X$, only by units. The idea of these equivalences is to leave the row space of the matrix invariant. If you multiply a row by $X$, you change the row space, since this row now generates a smaller ideal. You can check that the rows of that last matrix no longer span the first and third of your original generators, whereas the right-hand matrix in the first displayed equation that you got by basic row operations does still have all three generators in its row space.
Some more things to set straight: By "units in $\mathbb{Q}[X]^3$", you probably mean units in $\mathbb{Q}[X]$? You can turn $\mathbb{Q}[X]^3$ into a ring, too, with componentwise operations, but that doesn't enter into it when you consider $\mathbb{Q}[X]^3$ as a module over $\mathbb{Q}[X]$. Speaking of which, you didn't make explicit whether you're considering it as a module over $\mathbb{Q}[X]$ or over $\mathbb{Q}$. Since you wanted to multiply by $X$, I assumed the former; obviously in a module over $\mathbb{Q}$ multiplying by $X$ would be no less invalid. Lastly, your title seems confused: You're looking for a basis of a (sub)module in $\mathbb{Q}[X]^3$ over the ring $\mathbb{Q}[X]$ of polynomials.
