The Subring of Polynomials with Subring Coefficients

# The Subring of Polynomials with Subring Coefficients

Recall from the Subrings and Ring Extensions page that if $(R, +, *)$ is a ring, $S \subseteq R$, and $S$ forms a ring under the same $+$ and $*$ defined on $R$ then the ring $(S, +, *)$ is said to be a subring of $(R, +, *)$ and $(R, +, *)$ is said to be a ring extension of $(S, +, *)$.

We also saw that if $(R, +, *)$ is a ring where $0$ is the identity of $+$ and $1$ is the identity of $*$ then we can check if a subset $S \subseteq R$ is a ring by verifying the following ring axioms:

• $S$ is closed under $+$.
• For every element $a \in S$ there exists a $-a \in S$ such that $a + (-a) = 0$ and $(-a) + a = 0$.
• $S$ is closed under $*$.
• $1 \in S$.

Now recall from The Ring of Polynomials with Ring Coefficients page that if $(R, +_1, *_1)$ is a ring then $R[x]$ is the set of all polynomials whose coefficients are from $R$, that is, for $a_0, a_1, ..., a_n \in R$:

(1)
\begin{align} \quad a_0 + a_1x + a_2x^2 + ... + a_nx^n \in R[x] \end{align}

Now suppose that $(S, +_1, *_1)$ is a subring of $(R, +_1, *_1)$. It would be nice to know whether or not $S[x]$, the set of all polynomials whose coefficients are from $S \subseteq R$ is a ring under $+$ and $*$ defined on $R[x]$. As we will see in the following theorem, $S[x]$ is indeed a ring!

Since $S \subseteq R$ we have that the set of all polynomials with coefficients from $S$ is a subset of the set of all polynomials with coefficients from $R$, that is, $S[x] \subseteq R[x]$. Therefore we only need to check the four ring axioms above to determine whether $S[x]$ is a subring.

We first consider if $S[x]$ is closed under $+$. Let $p, q \in S[x]$ where the coefficients of $p$ are $a_i$s and the coefficients of $q$ are $b_i$s. Each term of the sum $p + q$ is of the form $(a_i +_1 b_i)$ which is contained in $S$ since $(S, +_1, *_1)$ is a ring. Therefore $(p + q) \in S[x]$ so $S[x]$ is closed under $+$.

Now let $0$ be the identity of $+_1$ on $S$. For each element $p \in S[x]$ we have that $-p \in S[x]$ is the polynomial whose coefficients are the inverses of the corresponding coefficients in $p$ with respect to $+_1$. So the coefficients of $p + (-p)$ are of the form $a_i +_1 (-a_i) = 0 \in S[x]$ and $(-p) + p = (-a_i) +_1 a_i = 0 \in S[x]$. So for every element $p \in S[x]$ there exists a $-p \in S[x]$ such that $p + (-p) = 0$.

Now consider the product $p * q$. The coefficients of $p * q$ with be be the product $*_1$ of some collection of coefficients from $S$. Since $S$ is closed under $*_1$ we see that the coefficients of $p * q$ will be contained in $S$. Therefore $(p*q) \in S[x]$ so $S[x]$ is closed under $*$.

Lastly, if $1$ is the multiplicative identity for the subring $S$ then the polynomial $1(x) = 1$ is the multiplicative identity for $S[x]$.

Therefore $(S[x], +, *)$ is a subring of $(R[x], +, *)$.