#### Exercise 4.4.1

Let $\mathbb Q$ be the field of rational numbers. Determine which of the following subsets of $\mathbb Q[x]$ are ideals. When the set is an ideal, find its monic generator.

(a) all $f$ of even degree;

(b) all $f$ of degree $\geq5$;

(c) all $f$ such that $f(0)=0$;

(d) all $f$ such that $f(2)=f(4)=0$;

(e) all $f$ in the range of the linear operator $T$ defined by

$$T\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1}.$$Solution:

(a) This is not an ideal. Let $f(x)$ be any polynomial of even degree. Let $g(x)=x$. Then $\deg(fg)=\deg(f)+1$ which is odd and is therefore the set of polynomials of even degree does not satisfy the necessary property that $fg$ is in the set whenever $f$ is in the set.

(b) This is not an ideal. Let $f(x)=x^5$ and $g(x)=-x^5+x^4$. Then $f$ and $g$ are in the set and therefore $f+g$ must be in the set if it is an ideal. But $f(x)+g(x)=x^4$ has degree equal to four and therefore is not in the set. In other words the set is not closed with respect to addition and therefore is not even a subspace.

(c) This is an ideal. Let $I=\{f\in F[x]\mid f(0)=0\}$. If $f(0)=0$ and $g(0)=0$ then $$(cf+g)(0)=cf(0)+g(0)=0$$ thus $I$ is a subspace of $F[x]$. Now suppose $f\in I$ and $g\in F[x]$. Then $$(fg)(0)=f(0)g(0)=0\cdot g(0)=0.$$ Thus $fg\in I$. Thus $I$ is an ideal. Let $f(x)=a_0+a_1x+\cdots+a_nx^n$. Then $f(0)=a_0$. Thus if $f\in I$ then necessarily $a_0=0$. Let $g(x)=x$. Then $g\in I$ and $$f(x)=a_1x+a_2x^2+\cdots a_nx^n=x(a_1+a_2x+\cdots a_{n}x^{n-1})\in I.$$ Thus $g$ generates $I$.

(d) This is an ideal. Let $I=\{f\in F[x]\mid f(2)=f(4)=0\}$. If $f(2)=f(4)=0$ and $g(2)=g(4)=0$ then $(cf+g)(2)=cf(2)+g(2)=0$ and $(cf+g)(4)=cf(4)+g(4)=0$ thus $I$ is a subspace of $F[x]$. Now suppose $f\in I$ and $g\in F[x]$. Then $$(fg)(2)=f(2)g(2)=0\cdot g(2)=0$$ and $$(fg)(4)=f(4)g(4)=0\cdot g(4)=0.$$ Thus $fg\in I$. Thus $I$ is an ideal. Let $g(x)=(x-2)(x-4)$. We claim $g$ generates $I$. Let $f(x)\in I$. Then since $f(2)=0$, by Corollary 1 page 128 it follows that $f(x)=(x-2)q(x)$ for some $q(x)\in F[x]$. Now since $f(4)=2\cdot q(4)=0$ it follows that $q(4)=0$ Thus there is a $p(x)$ such that $q(x)=(x-4)p(x)$. Thus $$f(x)=(x-2)(x-4)p(x)=g(x)p(x)$$ and therefore $f(x)$ is in the ideal generated by $g(x)$.

(e) This is an ideal. In fact it is the same ideal as in part (c). Note that $T(f)$ has no constant term, thus $T(f)(0)=0$ $\forall$ $f\in F[x]$. Now let $f(x)=a_1x+a_2x^2+\cdots+a_nx^n$. Let $$g(x)=a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1}.$$ Then $f=T(g)$. Thus $f$ is in the image of $T$. Thus any polynomial with zero constant term is in the image of $T$. Thus it is exactly the same ideal is in part (c).

#### Exercise 4.4.2

Find the g.c.d. of each of the following paris of polynomials

(a) $2x^5-x^3-3x^2-6x+4$, $x^4+x^3-x^2-2x-2$;

(b) $3x^4+8x^2-3$, $x^3+2x^2+3x+6$;

(c) $x^4-2x^3-2x^2-2x-3$, $x^3+6x^2+7x+1$.

Solution:

This question on its face seems to depend on what the base field is. I'm going to assume it is $\mathbb C$.

(a) Let $f(x)=2x^5-x^3-3x^2-6x+4$ and $g(x)=x^4+x^3-x^2-2x-2$. I have a feeling there's a mistake and they really meant $2x^5-x^3-3x^2-6x+6$ because then both polynomials are divisible by $x^2-2$. But as it is, the g.c.d. of these two polynomials equals 1 - which does not seem that easy to prove using only what we know up to this point. If we knew about unique factorization in $\mathbb C[x]$ we could simply factor $g(x)$ into linear factors and check none of the roots of $g$ are roots of $f$. In fact the roots of $g(x)$ are $\pm\sqrt2$ and $\frac{-1\pm\sqrt{2}i}{2}$.

But we can't use that argument yet. Instead we are probably expected to argue using the comments on page 132 right after the proof of Theorem 7, commonly known as the ``euclidean algorithm''.

Let $I=\langle f(x),g(x)\rangle$ be the ideal in $\mathbb C[x]$ generated by $f$ and $g$. Then

$$2x^5-x^3-3x^2-6x+4=(x^4+x^3-x^2-2x-2)(2x-2)+(3x^3-x^2-6x).$$Therefore $3x^3-x^2-6x=f(x)-(2x-2)g(x)\in I$.

$$x^4+x^3-x^2-2x-2=(3x^3-x^2-6x)\left({\textstyle\frac13}x+{\textstyle\frac49}\right)+\left({\textstyle\frac{13}{9}}x^2+{\textstyle\frac23}x-2\right).$$Therefore ${\textstyle\frac{13}{9}}x^2+{\textstyle\frac23}x-2\in I$.

$$3x^3-x^2-6x=\left({\textstyle\frac{13}{9}}x^2+{\textstyle\frac23}x-2\right)\left({\textstyle\frac{27}{13}}x-{\textstyle\frac{9\cdot31}{13^2}}\right) + \left({\textstyle\frac{-2\cdot3^2\cdot7}{13^2}}x-{\textstyle\frac{2\cdot3^2\cdot31}{13^2}}\right).$$Therefore $\frac{-2\cdot3^2\cdot7}{13^2}x-\frac{2\cdot3^2\cdot31}{13^2}\in I$.$${\textstyle\frac{13}{9}}x^2+{\textstyle\frac23}x-2 = \left({\textstyle\frac{-2\cdot3^2\cdot7}{13^2}}x-{\textstyle\frac{2\cdot3^2\cdot31}{13^2}}\right)\left({\textstyle\frac{-13^3}{2\cdot3^4\cdot7}}x+{\textstyle\frac{13^219^2}{2\cdot3^47^2}}\right)+\left({\textstyle\frac{13^261}{3^27^2}}\right).$$Therefore ${\textstyle\frac{13^261}{3^27^2}}\in I$. And $\frac{3^27^2}{13^261}\cdot\frac{13^261}{3^27^2}=1$ therefore $1\in I$. Therefore $I=\mathbb C[x]$. Therefore the g.c.d. of $f(x)$ and $g(x)$ equals 1: gcd$(f,g)=1$.

(b) Let $f(x)=3x^4+8x^2-3$ and $g(x)=x^3+2x^2+3x+6$. Let $I=\langle f(x),g(x)\rangle$. We have

$$3x^4+8x^2-3=(x^3+2x^2+3x+6)(3x-6)+(11x^2+33).$$Thus $11x^2+33\in I$.

Thus $x^2+3\in I$.Now

$$x^3+2x^2+3x+6=(x^2+3)(x+2).$$Thus $x^2+3$ divides both $f(x)$ and $g(x)$. In particular

$$f(x)=(3x^2-1)(x^2+3)$$$$g(x)=(x+2)(x^2+3)$$

and therefore it follows that$$\langle f(x),g(x)\rangle=\langle x^2+3\rangle.$$(c) As in part (a) I have a feeling there's a typo and they really meant $x^3+6x^2+7x+2$ because then both are divisible by $x+1$. But as it is, $x^3+6x^2+7x+1$ is irreducible and the calculations are even worse than part (a). As I did in part (a), I'll solve this in both the way they actually stated it and the was I think they intended it. First the way they stated it.

Let $f(x)=x^4-2x^3-2x^2-2x-3$ and $g(x)=x^3+6x^2+7x+1$ and $I=\langle f(x),g(x)\rangle$.

$$x^4-2x^3-2x^2-2x-3=(x^3+6x^2+7x+1)(x-8)+(33x^2+53x+5)$$Thus $33x^2+53x+5\in I$.

$$x^3+6x^2+7x+1=(33x^2+53x+5)\left({\textstyle\frac{1}{33}}x+{\textstyle\frac{145}{1089\cdot33^2}}\right)+\left({\textstyle\frac{-227}{1089}}x+{\textstyle\frac{364}{1089}}\right).$$Thus

$${\textstyle\frac{-227}{1089}}x+{\textstyle\frac{364}{1089}}\in I$$And finally

$$33x^2+53x+5=\left({\textstyle\frac{-227}{1089}}x+{\textstyle\frac{364}{1089}}\right)\left({\textstyle\frac{-1089\cdot33}{227}}x-{\textstyle\frac{26182827}{51529}}\right)+\left({\textstyle\frac{9009297}{51529}}\right).$$Thus $\frac{9009297}{51529}\in I$ and it follows that $1\in I$ and that gcd$(f(x),g(x))=1$.

#### Exercise 4.4.3

Let $A$ be an $n\times n$ matrix over a field $F$. Show that the set of all polynomials $f$ in $F[x]$ such that $f(A)=0$ is an ideal.

Solution: Let $I$ be the set $\{f\in F[x]\mid f(A)=0\}$. Let $f,g\in I$ and $c\in F$. Then $$(cf+g)(A)=cf(A)+g(A)=c\cdot0+0=0.$$ Thus $I$ is a subspace of $F[x]$ as a vector space over $F$. Now suppose $f\in I$ and $g\in F[x]$. Then $$(gf)(A)=g(A)f(A)=g(A)\cdot0=0.$$ Thus $I$ has the required property to be an ideal.

#### Exercise 4.4.4

Let $F$ be a subfield of the complex numbers, and let

$$A=\left[\begin{array}{cc}1&-2\\0&3\end{array}\right].$$Find the monic generator of the ideal of all polynomials $f$ in $F[x]$ such that $f(A)=0$.

Solution: Let $I=\{f(x)\in F[x]\mid f(A)=0\}$. We know from the previous exercise that $I$ is an ideal. It's clear that $f(A)\not=0$ if $\deg(f)\leq 1$. Thus if we find any $f(x)\in I$ such that $\deg(f)=2$ then $f$ must be a generator. Let $f(x)=x^2-4x+3$. Then

$$f(A)=A^2-4A+3$$$$=\left[\begin{array}{cc}1&-2\\0&3\end{array}\right]\cdot\left[\begin{array}{cc}1&-2\\0&3\end{array}\right]-4\left[\begin{array}{cc}1&-2\\0&3\end{array}\right]+3\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$$$=\left[\begin{array}{cc}1&-8\\0&9\end{array}\right]-\left[\begin{array}{cc}4&-8\\0&12\end{array}\right]+\left[\begin{array}{cc}3&0\\0&3\end{array}\right]=\left[\begin{array}{cc}0&0\\0&0\end{array}\right].$$

#### Exercise 4.4.5

Let $F$ be a field. Show that the intersection of any number of ideals in $F[x]$ is an ideal.

Solution: Let $A$ be a set. Let $I_{\alpha}$ be an ideal in $F[x]$ for each $\alpha\in A$. Let $I=\cap_{\alpha\in A}I_{\alpha}$. By Theorem 2, page 36 (sec. 2.2), $I$ is a subspace of $F[x]$. We must just show $I$ satisfies the extra condition required to be an ideal. Specifically, let $g(x)\in F[x]$ and $f(x)\in I$. We must show $g(x)f(x)\in I$. Since $f(x)\in I$, it follows that $f(x)\in I_{\alpha}$ $\forall$ $\alpha\in A$. Since $I_{\alpha}$ is an ideal $\forall\alpha\in A$, it follows that $g(x)f(x)\in I_{\alpha}$ $\forall$ $\alpha\in A$. Thus $f(x)g(x)\in I_{\alpha}$ $\forall$ $\alpha\in A$. Thus $f(x)g(x)\in I$.

#### Exercise 4.4.6

Let $F$ be a field. Show that the ideal generated by a finite number of polynomials $f_1,\dots,f_n$ in $F[x]$ is the intersection of all ideals containing $f_1,\dots,f_n$.

Solution: Let $I=\langle f_1,\dots,f_n\rangle$ be the ideal generated by $\{f_1,\dots,f_n\}$. Let $J$ be any ideal containing $\{f_1,\dots,f_n\}$. Let $g\in I$. Then $$g=g_1f_1+\cdots+g_nf_n$$ for some $g_1,\dots,g_n$ in $F[x]$. Since $J$ is an ideal containing $\{f_1,\dots,f_n\}$ it must also contain $g$. Thus $g\in I$ $\Rightarrow$ $g\in J$. Thus $I\subseteq J$. Now by the definition of intersection, $J$ is contained in every ideal which contains $\{f_1,\dots,f_n\}$. Since $I$ is such an ideal, it follows that $J\subseteq I$. Thus we've shown $I\subseteq J$ and $J\subseteq I$. Thus $J=I$.

Exercise 4.4.7

Let $K$ be a subfield of a field $F$, and suppose $f$, $g$ are polynomials in $K[x]$. Let $M_K$ be the ideal generated by $f$ and $g$ in $K[x]$ and $M_F$ be the ideal they generate in $F[x]$. Show that $M_K$ and $M_F$ have the same monic generator.

Solution 1: We first note a general fact. Let $E$ be any field. Let $f,g\in E[x]$ such that gcd$(f,g)=k$. Let $h\in E[x]$ be monic. We claim gcd$(kf,kg)=kh$ in $E[x]$. Since gcd$(f,g)=k$, the ideal generated by $f$ and $g$ is the same as the ideal generated by $k$. In other words $fE[x]+gE[x]=kE[x]$. Thus $hfE[x]+hgE[x]=hkE[x]$. Since $h$ is monic by assumption and $k$ is monic by definition of gcd, it follows that $hk$ is monic - and therefore $hk$ satisifes the definition of gcd$(f,g)$ in $E[x]$.

Now let $h=\text{gcd}(f,g)$ in $K[x]$. So $h\in K[x]$ and $h=fa+gb$ for some $a,b\in K[x]$. And $\exists$ $u,v\in K[x]$ such that $f=uh$ and $g=vh$. Once you can write the three equations $h=fa+gb$, $f=uh$ and $g=vh$ it follows that the ideal in $K[x]$ generated by $f$ and $g$ is the same as the ideal generated by $h$. That's what it means for $h$ to be the g.c.d. of $f$ and $g$. But these three equalities also hold in $F[x]$ and consequently the ideal in $F[x]$ generated by $f$ and $g$ is the same as the one generated by $h$.

Solution 2: This might be easier if they had formalized what is discussed informally on page 132 after the proof of Theorem 7. What they are describing is the so-called ``eucidean algorithm''. If you have two polynomials $f$ and $g$ and say $\deg(f)\geq\deg(g)$. Then you divide $g$ into $f$ and take the remainder $r\in K[x]$. That remainder is also in the ideal generated $f$ and $g$. Then replace $f$ and $g$ with $g$ and $r$ and do the same thing, divide $r$ into $g$ and the remainder has degree less than $r$. Eventually you must arrive at a remainder equal to zero or of degree equal to zero (i.e. a scalar). If the remainder is zero then the previous remainder generates the ideal $\langle f,g\rangle$ in $K[x]$. If the remainder is a non-zero scalar then the ideal contains $1$ and therefore $\langle f,g\rangle=K[x]$. The euclidean algorithm allows us to always find the gcd in a finite number of steps.

Once we have the euclidean algorithm, all we need to do to solve this problem is to note that all operations in the euclidean algorithm happen in $K[x]$ and so if $K\subseteq F$ then in $F[x]$ the same operations will lead to the same generator which will therefore still live in the smaller ring $K[x]$.

From http://greggrant.org