Why is this J(all the linear combinations of two polynoimials in F[x]) an ideal of polynomial field F[x]?

Question

Suppose $a(x)$ and $b(x)$ are two non-zero polynomials in the polynomial field $F[x]$ have a gcd $d(x)$ can be expressed as a "linear combination": $$ d(x) = r(x)a(x) + s(x)b(x) $$ where $r(x)$ and $s(x)$ are in $F[x]$

Now if $J$ is the set of all the linear combinations:

$$ u(x)a(x) + v(x)b(x) $$

as $u(x)$ and $v(x)$ range over $F[x]$ then $J$ is an ideal of $F[x]$.

I don't understand this last statement "$J$ is an ideal of $F[x]$"

According to the definition of ideal:

"A nonempty subset $B$ of a ring $A$ is called an ideal of $A$ if $B$ is closed with respect to addition and negatives and $B$ absorbs products in $A$."

Now for the above problem how does this definition of ideal reasserts the idea that $J$ is an ideal?

Can anyone kindly help me find the answer?

Answer 1

Your definition of ideal is correct, so now we need to show that $J$ satisfies the definition. So we ask the following questions:

• Is $J$ closed under addition? That is, if $r_1(x)a(x)+s_1(x)b(x)$ and $r_2(x)a(x)+s_2(x)b(x)$ are in $J$, is their sum also in $J$?
• Is $J$ closed under multiplication by elements in $F[x]$? That is, if $p(x)\in F[x]$, is the product of $p(x)$ and $r(x)a(x)+s(x)b(x)$ in $J$?

The answer to both of these questions is 'yes,' and this is shown by remembering that the elements in $J$ are all possible linear combinations of the polynomials $a(x)$ and $b(x)$. See if you can put the pieces together.

Notice that we can take $p(x)=-1$ in the second question to show that $J$ is closed under negatives.