# polynomial ring

• October 23rd 2009, 11:19 AM
knguyen2005
polynomial ring
Let R be a ring, suppose that 2 non-zero functions f and g with:
f = a_0 + a_1x +...+ a_mx^m belongs to R[x] , a_m not equal 0 and deg(f) = m
g = b_0 + b_1x +....+ b_nx^n belongs to R[x], b_n not equal 0 and deg(g) = n

(a) Let R be and Integral Domain (ID). Show that f.g is nonzero and f.g has leading term is (a_m)(b_n)x^(m+n). Deduce that R[x] is a ID and deg(f.g) = deg(f) +deg(g). If R[x] is a ID, show that R is also a ID.
(b) Let R be a ID. Show that U(R[x]) = U(R)
(c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1?
(d) Show that U(Z_4) is strictly not subset of U(Z_4[x]). Why doesn't this contradict part (b)?
Note: Z_4 is set of integer mod 4

Ok, Part (a) is easy and straightforwards.
f.g is nonzero sine both f and g are nonzero.
Also, the highest power of f.g is (a_m)(b_n)x^(m+n) obtained when u multiplied 2 functions.
R[x] is an ID because either f = 0 or g = 0 (not both equal o at the same time)
deg(f.g) = m + n = deg(f) +deg(g)

Part (b), (c) and (d) I dont know how to begin with

Can someone show me how to do part b, c and d please?

Thank you
• October 23rd 2009, 03:07 PM
proscientia
I don’t think you’ve even done part (a) properly.

Recall that the coefficient of $x^k$ in $fg$ is

$\sum_{i\,=\,0}^ka_ib_{k-i}$

When $k=m+n,$ the sum is just $a_mb_n$ because $a_i=0$ for all $i>m$ and $b_{m+n-i}=0$ for all $i Hence the coefficient of $x^{m+n}$ in $fg$ is $a_mb_n.$ And since $a_m\ne0,\ b_n\ne0$ and $R$ is an integral domain, $a_mb_n\ne0$ and so $fg$ is nonzero.

Do you understand this so far?
• October 23rd 2009, 09:12 PM
aliceinwonderland
Quote:

Originally Posted by knguyen2005
(a) .....
If R[x] is a ID, show that R is also a ID.

Let $f(x)=a_0 + a_1x + \cdots +a_mx^m$ and $g(x)=b_0 + b_1x + \cdots + b_nx^n$ be two nonzero polynomials in R[x]. Since R[x] is an integral domain, f(x)g(x) is nonzero and we have a non-zero leading term $a_kb_lx^{k+l}$. This follows that $a_k \neq 0$ and $b_l \neq 0$. Since R[x] is an integral domain, we obtain $a_kb_l \neq 0$. Thus R is an integral domain.
Think about this argument $((a_k \neq 0 \wedge b_l \neq 0) \Rightarrow a_kb_l \neq 0) \Leftrightarrow (a_kb_l = 0 \Rightarrow (a_k = 0 \vee b_l=0))$.

Quote:

(b) Let R be a ID. Show that U(R[x]) = U(R)
A polynomial f(x) of degree m times a polynomial g(x) of degree n is a polynomial f(x)g(x) of degree m+n in an integral domain R[x]. Suppose f(x) is a polynomial whose degree is 1 or greater. Then no polynomial g(x) exists such that f(x)g(x)=1, which is a polynomial of degree 0. This forces that units of R[x] only exists with a degree zero. Thus U(R[x])=U(R).

Quote:

(c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1?
Think about the Maclaurin series of 1/(1-a) and use a nilpotency of a.

$(1-a)(1+a+ \cdots +a^{n-1}) = 1$, where $a^n=0$

Quote:

(d) Show that U(Z_4) is strictly (not?) subset of U(Z_4[x]). Why doesn't this contradict part (b)?
Note: Z_4 is set of integer mod 4
$U(\mathbb{Z}_4) = \{1,3\}$. They are coprime to 4. Meanwhile $U(\mathbb{Z}_4[x])$ has additional units including $U(\mathbb{Z}_4)$. For instance, 1 + [2]x is a unit in $U(\mathbb{Z}_4[x])$. We know that $\mathbb{Z}_4$ is not an integral domain. This implies that if R is not an integral domain, $U(R) \subset U(R[x])$.
• October 24th 2009, 04:13 AM
knguyen2005
Thank you both proscientia and aliceinwonderland
I already understood and know how to do it now

Thanks again for spending your time on this question