# No linear polynomial combination in an ideal

tough (*) question from an old examination paper:

$I=4R+2xR+x^{2}R$ is an ideal of $R= \mathbf{Z}[x]$. Show that there are no $U,V$ in $R$ , so that $I=UR+VR$ holds.

Hardmath, Thank you! You showed me thouroughly how wrong I was. However I still can not figure out how you treat the case $2x$ and the case $x^{2}$ by setting the polynomials to certain values. Help is greatly appreciated.

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Hint: I would start with the "constant" part of ideal I, namely that it contains $4R$. How does this constrain your $U,V$ ? Obviously some combination of $U,V$ must give 4 (but no combination gives a proper divisor of 4). Similarly there must be combinations that give $2x$ and $x^2$. I suspect that by the time you assemble these facts, you'll be close to using a degree argument that $U,V$ cannot exist. – hardmath Nov 24 '11 at 14:47
Hardmath, Thank you! I'm not sure whether I have fully understood your argument or not. So if U,V gives 4, e.g. U would be a polynomial with constant 1 and V would be a polynomial with constant 3. Then at the same time the x coefficients must add up to 2. So f.e. my example would have $u_{1}=1$ and $v_{1}=1$ . Then for the quadratic term $u_{2} = 0$ and $v_{2} = 1$ I'm not quite sure what to do with this so far, though. – PumaDAce Nov 24 '11 at 16:36
I've supplied the details in an informal sketch. Let me know if you'd like elaboration of a few points, or simply a more formal proof. – hardmath Nov 29 '11 at 17:24
Hardmath, thank you so much. Especially since there are no solutions available to these old examination papers, this will be a keeper for many to come! – PumaDAce Dec 1 '11 at 1:58

This is meant to correct a misimpression left by my hint, and not (yet) to serve as a complete answer.

Given the ring of polynomials $R = \mathbb{Z}[x]$ over the integers, we are asked to show the ideal $I = (x^2,2x,4)$, which is evidently generated by three elements, cannot actually be generated by any two polynomials $U,V$ in $R$.

As you might guess this is a special case of a more general theorem, but we are likely expected to limit ourselves to elementary methods in proving the special case.

The first trick to consider here is passing to the quotient ring by taking polynomials modulo the prime ideal $(x)$, which puts us back in familiar territory, the ring $\mathbb{Z}$.

Another way to view that homomorphism of rings is "evaluation at zero", i.e. setting $x=0$ in every polynomial. Now the statement that $U,V$ generate 4 in $R$ becomes an easier idea to work with. That is from polynomials $p_0,q_0$ that give us in ring $R$:

$$p_0*U + q_0*V = 4$$

we get an inference about the constant terms of $U,V$, by applying the homomorphism:

$$p_0(0)*U(0) + q_0(0)*V(0) = 4$$

in the ordinary ring of integers $\mathbb{Z}$. Moreover as outlined in my comment, no combination of $U(0),V(0)$ can give a proper divisor of 4, else the image of $I$ under the ring homomorphism would contain more than the ideal $4\mathbb{Z}$.

Thus your guess that $U,V$ have constant terms 1 and 3, respectively, is incorrect. What we can say is that the greatest common divisor of $U(0),V(0)$ is $4$.

We can go further with the analysis and show that without loss of generality: $$U = U^*x^2 + 2mx + 4$$ $$V = V^*x^2 + (4n+2)x$$ Once we have this, we can show that $x^2$ is not generated by $U,V$.

The argument goes like this. Suppose $pU + qV = x^2$. Since the constant term of $V$ is zero, $p$ must have a factor of $x$ to avoid contributing a nonzero constant to the combination. But then $pU$ has a coefficient of $x$ that is a multiple of $4$, and to avoid $qV$ contributing an uncancelled (mod $4$) coefficient of $x$, the constant term of $q$ must be even.

Putting these facts together tells us that the $x^2$ coefficient in $pU + qV$ is even, contradicting our assumption that it equals $x^2$. For $p$ contains a factor of $x$, and so $pU$ must make the even contribution that is a multiple of $2mx^2$, from the lowest degree term in $p$ times the $x$ term in $U$. And the contribution of $qV$ to the $x^2$ term consists of the even constant term of $q$ times the $x^2$ term of $V$ plus the $x$ term of $q$ times the even term $(4n+2)x$ of $V$. QED

It remains to back up and explain why without loss of generality $U,V$ may be taken to have the forms shown above. First of all, since $U,V$ belong to ideal $(x^2,2x,4)$, it is clear that their constant terms will be multiples of $4$ and that their $x$ terms will be even.

But more than this, as previously explained, $4$ is the greatest common divisor of the constant terms $U(0),V(0)$. Thus the steps of the Euclidean algorithm in $\mathbb{Z}$ lead to expressing $4$ as an integer combination of $U(0),V(0)$.

Recalling how the Euclidean algorithm works in $\mathbb{Z}$, apply those steps of adding an integer multiple of one polynomial $U,V$ to the other, until finally we reach a pair of polynomials which have constant terms $4$ and zero respectively.

Note that the ideal $(U,V)$ is equal to the ideal $(U+kV,V)$. For the latter is surely contained in the former, and it is easy to see that $U,V$ are each contained in the latter.

So repeatedly apply those steps of the integer Euclidean algorithm will lead to the promised pair of polynomials, which we take the liberty of relabelling so that $U(0) = 4$ and $V(0) = 0$.

Finally we argue that the $x$ term of $V$ must be "oddly even", i.e. congruent to $2$ mod $4$. This is a consequence of expressing $2x$ as a combination of $U,V$. For again the multiple of $U$ must have a factor of $x$ to avoid contributing a nonzero constant term, and thus said multiple of $U$ will have an $x$ term whose coefficient is a multiple of $4$. To get $2x$ as a combination therefore requires the $x$ term of $V$ to be of the form $(4n+2)x$.

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