# 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 $\mathbb{Q}[X]$.

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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.
Sorry joriki, I've tried to change the title to reflect that I'm looking at a submodule in $\mathbb{Q}[X]^3$ over the ring $\mathbb{Q}[X]$. If I can't multiply by $X$, do I just stop and take $(X,X,X^2)$ and $(1,X,X^2-3)$ as a basis? – KARA Oct 27 '11 at 16:23
@KARA: The title is OK now. It still says "units in $\mathbb{Q}[X]^3$". Regarding the basis: Yes. Those two elements are linearly independent, since there are no elements of $\mathbb{Q}[X]$ that would make both $aX+b1$ and $aX+bX$ come out to zero. – joriki Oct 27 '11 at 16:42
Ok, thanks joriki. I fixed the units phrase. So in general, I can swap rows, add and subtract them, and multiply them by units in $\mathbb{Q}[X]$. Is there a way to know when it's sufficient to stop? I'm used to being able to put a matrix in upper triangular form when working over a field. With a ring that seems not to be the case. – KARA Oct 27 '11 at 18:25
@KARA: You can always bring a matrix over a Euclidean ring into upper triangular form, since you can reduce the degrees of the entries of higher degree using the Euclidean algorithm. This is not true for general rings. In the present case, $\mathbb Q[X]$ is a polynomial ring over a field and therefore Euclidean. You can bring your matrix into upper triangular form by subtracting $X$ times the second row from the first and then swapping the rows. It's OK to do $a\to a+Xb$, just not $a\to Xa$ if $X$ isn't a unit. One is invertible, the other isn't. – joriki Oct 27 '11 at 18:46