Topics in Algebra, Chapter 3.11, Part 1

This section covers section 3.11 ("Polynomial Rings over Commutative Rings"). This page is split into parts so as to not get excessively long.
**See also**: [part 2](/journal/topics-in-algebra-chapter-3-section-11b) (3.11.9 through 3.11.15).
### Topics covered: 3.11
* **Definition**: Let $R$ be a commutative ring with a unit element. $R[x]$ is defined, as in section 3.9, as the set of all symbols $a_0+a_1 x+\cdots+a_m x^m$ where $m$ is a non-negative integer and $a_0,\ldots,a_m\in R$. The standard addition and multiplication make $R[x]$ a ring. It will be shown in exercise 3.11.1 that $R[x]$ so defined is a commutative ring with unit element.
* **Definition**: Let $R$ be a commutative ring with unit element. By problem 3.11.1, $R[x]$ is again a commutative ring with unit element. Therefore we may recursively define $R[x_1,\ldots,x_n]$ as $R[x_1,\ldots,x_{n-1}][x_n]$.
* **Lemma 3.11.1**: If $R$ is an integral domain, then $R[x]$ is an integral domain.
* **Corollary**: If $R$ is an integral domain, then $R[x_1,\ldots,x_n]$ is an integral domain for any positive integer $n$.
* **Definition**: Let $R$ be an integral domain. $R$ is a **unique factorization domain** (UFD) if (1) non-zero $r\in R$ is either a unit or expressible as a product of a finite number of irreducible elements of $R$, (2) the decomposition just mentioned is unique up to re-ordering and multiplication by units.
* **Lemma 3.11.2**: If $R$ is a UFD and $a,b\in R$, then $(a,b)\in R$, i.e. their gcd exists in $R$. Additionally, if $(a,b)=1$ then $a\mid bc$ implies that $a\mid c$.
* **Corollary**: If $R$ is a UFD and $a\in R$ is irreducible, then $a\mid bc$ implies $a\mid b$ or $a\mid c$.
* **Lemma 3.11.3**: If $R$ is a UFD and $f,g\in R[x]$ are primitive, then $fg\in R[x]$ is primitive.
* **Lemma 3.11.4 (Gauss Lemma)**: Let $R$ be a UFD and hence an integral domain. By lemma 3.11.1 and theorem 3.6.1, $R[x]$ embeds in a field of fractions $F[x]$. If $f\in R[x]$ is primitive in $R[x]$, then it is irreducible in $R[x]$ if and only if it is irreducible in $F[x]$.
* **Lemma 3.11.5**: Let $R$ be a UFD and let $p\in R[x]$ be primitive. Then $f$ may be factored uniquely into irreducible elements in $R[x]$.
* **Theorem 3.11.1**: If $R$ is a UFD, then so is $R[x]$.
* **Corollary**: If $R$ is a UFD, then so is $R[x_1,\ldots,x_n]$ for any positive integer $n$.
* **Corollary**: If $F$ is a field, then $F[x_1,\ldots,x_n]$ is a UFD.
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
### Herstein 3.11.1: Let $R$ be a commutative, unital ring. Prove that $R[x]$ is a commutative unital ring.
This problem justifies the recursive definition of $R[x_1,\ldots,x_n]$. It is trivial but tedious (and notationally cumbersome) to show that $R[x]$ is a commutative ring, so I will take it for granted. The unit element is simply the constant polynomial $f\in R[x]$ given by $f(x)=1_R$ for all $x$.
### Herstein 3.11.2: Prove that $R[x_1,\ldots,x_n]\cong R[x_{\sigma(1)},\ldots,x_{\sigma(n)}]$ where $\sigma\in S_n$ permutes the indices.
What follows is a sketch of a proof. First consider a simple illustration using $R[x,y]$ and $R[y,x]$. It is easy to see that the elements of the former all have the form
$$\sum a_{ij}x^i y^j$$
while elements of the latter have the form
$$\sum a_{ij}y^i x^j.$$
Defining these polynomials as "formal symbols" as we do, it is not *a priori* obvious that monomials $r x^i y^j$ and $r y^j x^i$, with $r\in R$, are in any sense equal. Still, it is tempting to simply move the indeterminates around.
In $R[x_1,\ldots,x_n]$, the above comments suggest the mapping
$$\phi:R[x_1,\ldots,x_n]\to R[x_{\sigma(1)},\ldots,x_{\sigma(n)}]$$
given by
$$\phi\left(\sum a_{i_1\cdots i_n} x_1^{i_1}\cdots x_n^{i_n}\right)=\sum a_{i_1\cdots i_n} x_{\sigma(1)}^{i_{\sigma(1)}}\cdots x_{\sigma(n)}^{i_{\sigma(n)}}.$$
Here, the coefficient $a$ on each monomial remains the same while the positions of the indeterminates are permuted. For simplicity of notation, we can imagine each of $i_1,\ldots,i_n$ ranging from zero to infinity, although we note that only finitely many of the numbers in the $a$ array can be non-zero (a polynomial has finite degree) so there are no awkward questions of convergence, etc.
We assert that $\phi$ is an isomorphism, which is easy to believe. The only non-trivial thing to check is that $\phi$ respects ring multiplication. With correct bookkeeping, one sees that it does, but the details will be skipped here.
### Herstein 3.11.3: Let $R$ be an integral domain and let $f,g\in R[x]$. Prove that ${\rm deg}(fg)={\rm deg}(f)+{\rm deg}(g)$.
Let $f=r_0+r_1 x+\cdots+r_n x^n$ and let $g=s_0+s_1 x+\cdots+s_m x^m$, where all the coefficients belong to $R$ and $r_n,s_m\ne 0$. Then the degree of $f$ is $n$ and the degree of $g$ is $m$. The term of largest power in $fg$ is $r_n s_m x^{m+n}$, and its coefficient is $r_n s_m\ne 0$ because $R$ is an integral domain. Hence the degree of $fg$ is $m+n$.
### Herstein 3.11.4: Let $R$ be a unital integral domain and let $u\in R[x]$ be a unit. Show that $u$ is also a unit in $R$.
Let $v\in R[x]$ be such that $uv=1\in R[x]$. We have by problem 3.11.3 that $0={\rm deg}(uv)={\rm deg}(u)+{\rm deg}(v)$, which forces both $u$ and $v$ to have degree zero, i.e. to be elements of $R$. But then $uv=1\in R$ so that $u\in R$ is a unit. This is a slight abuse of notation, but the idea is clear that $u\in R[x]$ is a constant polynomial which we identify with its lone coefficient belonging to $R$.
### Herstein 3.11.5: An element $x$ in a ring is *nilpotent* if there exists positive $n\in\mathbb{Z}$ such that $x^n=0$. Let $R$ be a commutative ring with no non-zero nilpotent elements and let $f\in R[x]$ be a zero divisor given by $f(x)=a_0+a_1 x+\cdots+a_m x^m$. Prove there exists non-zero $b\in R$ such that $ba_0=ba_1=\cdots=ba_m=0$.
Because $f$ is a zero divisor, there exists $g\in R[x]$ such that $fg=0$. Then if $g(x)=b_0+b_1 x+\cdots+b_n x^n$, we have the product
$$(fg)(x)=\sum_{k=0}^{m+n}c_k x^k$$
where $c_k=a_k b_0+a_{k-1}b_1+\cdots+a_0 b_k$. Now this polynomial is identically zero, so each $c_k=0$. For instance, we see that $c_0=a_0 b_0=0$ and that $c_1=a_0 b_1+a_1 b_0=0$. Because $a_0 b_0=0$, we can try multiplying the second equation by $b_0$, in which case we see
$$0=a_0 b_0 b_1+a_1 b_0^2=a_1 b_0^2.$$
This suggests an induction: we claim that $b_0^{\ell+1}a_\ell=0$ for all $\ell< k$ and then we consider
$$0=b_0^k c_{k+1}=b_0^{k+1}a_k+b_0^k a_{k-1}b_1+b_0^k a_{k-2}b_2+\cdots+b_0^k a_0 b_k.$$
By the induction hypothesis, all terms vanish except the first, leaving
$$0=b_0^{k+1}a_k$$
which completes the induction.
Returning to the problem at hand, we see that $b=b_0^{m+1}$ is such that
$$ba_0=ba_1=\cdots=ba_m=0.$$
If $b_0\ne 0$ then we are done, because $b=b_0^{m+1}\ne 0$ by the prohibition on nilpotent elements. The case of $b_0=0$ is somewhat trivial. If $b_0=0$, we may rewrite $g$ as $g(x)=x^r h(x)$ where the degree-zero coefficient of $h(x)$ is non-zero, and then the above analysis goes through by using the coefficients of $h$ rather than those of $g$.
### Herstein 3.11.6*: Let $R$ be a commutative ring and let $f\in R[x]$ be a zero divisor given by $f(x)=a_0+a_1 x+\cdots+a_m x^m$. Prove there exists non-zero $b\in R$ such that $ba_0=ba_1=\cdots=ba_m=0$. This is the same as problem 3.11.5 without the nilpotency condition.
Here is a clever and subtle proof of this result due to [this math.stackexchange.com post](http://math.stackexchange.com/a/83171/37832).
Let non-zero $g\in R[x]$ be of minimal degree such that $fg=0$ and write $g=b_0+b_1 x+\cdots+b_n x^n$. Suppose that there does not exist any such $b$ having $ba_0=\cdots=ba_m=0$.
If all of the coefficients $a_i$, $i=0,\ldots,m$ annihilate $g$, having $a_i g=0$, then we would certainly have
$$a_i b_n=0$$
for every $i$. However, this would lead to
$$b_n f=b_n a_0+\cdots+b_n a_m x^m=0,$$
contradicting our supposition that no such $b=b_n$ exists.
Therefore if we accept that no such $b$ exists, then some coefficient fails to annihilate $g$. In particular, there is a largest index $i$ such that $a_i g\ne 0$. Now,
$$
\begin{multline*}
fg=\Big(a_0+\cdots+a_i x^i+\cdots+a_m x^m\Big)\Big(b_0+\cdots+b_n x^n\Big)\\
=\Big(a_0+\cdots+a_i x^i\Big)\Big(b_0+\cdots+b_n x^n\Big)=0
\end{multline*}
$$
because $a_j g=0$ for $j>i$. Expanding this product, it is seen that $a_i b_n=0$ so that ${\rm deg}(a_i g)< n$ because we have previously determined that $a_i g\ne 0$. We had supposed that $g$ was the non-zero polynomial of least degree which annihilated $f$, but now we are forced to conclude that $f\cdot(a_i g)=0$ as well. This contradiction proves our result.
The above proof does not exhibit any concrete element of $R$ which sends $f$ to zero. A [constructive proof](http://math.stackexchange.com/a/227787/37832) shows that every coefficient of $f$ must indeed annihilate $g$ (contrast with the second paragraph) and has us conclude that
$$b_n f=0.$$
### Herstein 3.11.7*: Let $R$ be a commutative, unital ring. Show that $f\in R[x]$ given by $f(x)=a_0+a_1 x+\cdots+a_n x^n$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1,\ldots,a_n$ are nilpotent in $R$.
There is intuition to be gained here from writing
$$\frac{1}{f}=\frac{1}{a_0(1+f')},$$
where $f'=(f-a_0)/a_0$, and then "Taylor expanding":
$$\frac{1}{f}=\frac{1}{a_0}\left(1-f'+f'^2-\cdots\right).$$
Roughly speaking, this should only make sense when (1) $a_0$ is invertible, and (2) the series terminates due to nilpotency. Now we seek to formalize this argument.
One way of doing so is to postulate this **lemma**: Let $S$ be a commutative, unital ring and let non-zero $n\in S$ be nilpotent with $n^k=0$. Then $1+n$ is a unit. **Proof**: Indeed, if we try the "Taylor expansion" as an inverse, we find
$$(1+n)(1-n+n^2-\cdots+(-1)^{k-1}n^{k-1})=1+(-1)^{k-1}n^k=1,$$
a telescoping sum. The series expansion for the inverse of $1+n$ may make sense in a formal context even if it does not terminate, but in our case we don't need to worry -- it terminates thanks to the nilpotency of $n$.$\quad\Box$ Furthermore, if $u\in S$ is any unit, we have $u+n=u(1+u^{-1}n)$ is again a unit because $u^{-1}n$ is nilpotent.
Now suppose that $a_0$ is a unit in $R$ and $a_1,\ldots,a_n$ are all nilpotent, say with nilpotency indices $k_i$ such that $a_i^{k_i}=0$. We have that $f_0=a_0$ is a unit, and, by the lemma, $f_1=a_0+a_1 x$ is again a unit because $a_1 x$ is nilpotent, $(a_1 x)^{k_1}=a_1^{k_1} x^{k_1}=0$. The lemma applies because $R[x]$ is a commutative, unital ring by exercise 3.11.1. Now repeating this stacking process $n-1$ more times, we conclude that $f=f_n$ is a unit.
On the other hand, suppose that $f$ were a unit and that
$$g(x)=b_0+b_1 x+\cdots+b_n x^n$$
were such that $fg=1$. Writing out the product explicitly, we have that $a_0 b_0=1$ so that $a_0$ must be a unit. At the highest degree terms, we find
$$a_n b_m=0$$
and
$$a_n b_{m-1}+a_{n-1}b_m=0.$$
Multiplying the second equation by $a_n$, the second term in the sum vanishes, and we are left with $a_n^2 b_{m-1}=0$. This suggests that ever higher powers of $a_n$ should annihilate each successive coefficient of $g$, counting downwards. Performing the relevant induction, we see that it is true and, as a result, $a_n^{m+1}$ annihilates every coefficient in $g$. Now
$$a_n^{m+1}=a_n^{m+1}(fg)=f\cdot(a_n^{m+1}g)=0$$
shows that $a_n$ is nilpotent.
Finally, we consider the polynomial
$$f-a_n x^n$$
which is a unit by the lemma, because $f$ is a unit and $a_n x^n$ was just shown to be nilpotent. Repeating the argument from above, we conclude that $a_{n-1}$ is nilpotent. By induction, the same may be said of each coefficient $a_i$ with $i>1$.
Some nudges toward this solution were provided by this [excellent site](http://www.mathreference.com/id-rad,nilu.html). There is also a very well known, elegant proof that unit $f$ implies nilpotent coefficients. It can be found in [this math.stackexchange.com post](http://math.stackexchange.com/a/82751/37832). However, it is clearly not the solution Herstein had in mind, because prime ideals have not been introduced at this point in the text. Also, to say that a prime ideal exists in $R$ is a non-trivial statement, [equivalent to the axiom of choice](http://en.wikipedia.org/wiki/Krull%27s_theorem).
### Herstein 3.11.8: Let $F$ be a field. Prove that $F[x_1,x_2]$ is not a principal ideal domain.
We try the most natural thing, which is to look at the ideal generated by $x_1$ and $x_2$. This hasn't been defined in the text, but we take a stab, writing
$$I=(x_1,x_2)=\{ax_1+bx_2\mid a,b\in F[x_1,x_2]\}.$$
It is trivial to see that $I$ is indeed an ideal of $F[x_1,x_2]$.
Let $d\in I$ be arbitrary. Because $F$ has no zero divisors and $d$ is built on top of $x_1$ and $x_2$, we must have ${\rm deg}(d)\ge 1$. Here we define the degree of a multivariate monomial as the sum of the powers on its indeterminates, and the degree of a multivariate polynomial as the largest out of the degrees of its constituent monomials. If $d$ were to generate the ideal, we would also have $d\mid x_1$ and $d\mid x_2$, which is only possible if $d$ is a constant, having degree zero. Therefore the ideal $I$ is not principal and $F[x_1,x_2]$ is not a principal ideal domain.