Given some goofy product like $(2x + \frac{63}{7})(5x + \frac{102}{5}) + 2$, is there some "hack" for moving the $2$ into one of the factors by eyeballing it without expanding the whole thing and refactoring?

@dc3rd what's the connection between "I have a function" and "I have a basis"

if they have nothing to do with one another, there's nothing to infer, of course

@user10478 if you have $x+2$, how do you put the $2$ anywhere else?

I mean you don't in that case.

Well, how is your situation different?

@Thorgott Well I was hoping I didn't have to be more explicit, but it may just be the case. So the question is from Insel's Linear Algebra:

Because my situation is $f(x)g(x) + 2$ where $f(x)g(x)$ has a constant term?

So I've done all the mechanical work before and I'm just reviewng my solution, and I noticed I didn't really show the existence

$x$ does not have a constant term

@Thorgott

ah, but that's a different scenario than what you just said

you don't have a function, you're supposed to show it exists

Ok, well I did computations and I got a value for $T(8,11)$. But is that sufficient to show it exists?

ok: so if you have $(x+1)x$ and you add $2$, so what do you expect to happen?

how are you able to do computations with something you don't even know exists

The same thing as $x + 2$ because $(x + 1)x$ has no non-trivial constant term :P

@Thorgott What's the issue?

Even in that case you can factor it more cleanly though.

Express $\sqrt 2 = p/q$ for integers $p,q$.

Computations

Contradiction.

Fortunately the solution I wrote out is exactly how another one online is written so I can post a screen capture of it. Once second @Thorgott

I already established linear independence above.

It's of course nicer and has real coefficients if you add $1$ to $(x + 2)x$.

@user10478 You're good at excuses. There is no hack.

You have to expand, simplify, and in general it will not factor.

Any quadratic factors, of course, if you allow complex numbers.

@Thorgott I think I see after looking at another question right after. So in the next question I get a situation where a set of vectors is not linearly independent and thus is not a basis of my domain. Whereas in this question the set of vectors is a basis for my domain and by a theorem a linear transformation is completely determined by its actions on a basis. So that's why the above is a linear transformation.

I guess I could also stick it into either term by dividing it by the other term to get a product of rationals.

@Karl I must've gravely offended the inaccessible cardinal theorists

@dc3rd right, the point, to be clear, is that not only is a linear transformation uniquely determined by its action on a basis, but also that for every possible action on a basis, there exists a linear transformation acting that way (and, due to the first part, this linear transformation is unique)

@Thorgott lol yes

That's actually interesting. It seems to imply that polynomials in $\mathbb{C}$ can be written as ratios of real polynomials.

Or not really, cause the polynomial is always real even when its factored form is complex, hmm.

the precise statement is that if $V,W$ are vector spaces and $v_1,...,v_n$ is a basis of $V$, then for every choice of vectors $w_1,...,w_n$ in $W$ there exists one and only one linear transformation $T\colon V\rightarrow W$ satisfying $f(v_i)=w_i$ for $i=1,..,n$

Yes I have the text in front me as well and remember this. I suppose writing it out gave me more clarity in understanding. THanks for the assistance.

Btw guys, are you aware of the slick proof of (weak nullstellensatz ==> strong nullstellensatz)?

How does it go?

Well, the writer of the proof called it the Rabinowitz trick (?), but so, say we have our ideal $\mathfrak a=(f_1,\dots,f_n)$ and $g\in I(Z(\mathfrak a))$, then $f_1,\dots,f_n$ and $x_{n+1}g-1$ have no common zeroes in $k^{n+1}$, and therefore by the weak nullstellensatz we get $1=p_1f_1+\dots p_nf_n + p_{n+1}(x_{n+1}g-1)$.

We can now consider the mapping $k[x_1,\dots,x_{n+1}]\to k(x_1,\dots,x_n)$ which sends $x_{n+1}$ to $1/g$. Then we find that $1=\sum_i p_i(x_1,\dots,x_n,1/g)f_i$, and by multiplying with some power of $g$ we find that this power lies in $\mathfrak a$

the rabinowitsch trick?

ah yes

Also, @Thorgott, I found a counter ex (on stack, that is) for this generator q I asked :o You can just consider $(x,y)^2\in k[x,y]$. And the argument there works for any graded ring where the base ring is a field (of fin dim)

yeah, right

kinda silly i didnt think of this

@TedShifrin Just missed you, I badly needed a nap. Howdy.

@Sha there's also a neat geometric interpretation of Rabinowitz's trick but I forgot all the details

@Fargle You're forgiven.

Phew.

This time!