Exercise Sheet 10

91) Let \(K\) be a field and \(p(x) \in K[x]\) a polynomial of degree \(m\). Prove that \(p(x)\) cannot have more than \(m\) zeros in \(K\) (counted with multiplicities).

Hint: Use the fact that \(K[x]\) is a factorial ring.

Firstly we recall:

\[\begin{align} p(a) = 0 &\ \Leftrightarrow (x-a) | p\\ \text{"$a$ has an $n$-fold zero"} &:\Leftrightarrow (x-a)^n | p \end{align}\]

By \(K[x]\) being a factorial ring, \(p\) can be written as:

\[\begin{gather*} p = u \prod_i p_i, \quad i \geq 0 \end{gather*}\]

We can partition this product as follows:

\[\begin{gather*} p = u \prod_i (x-a_i)^{ni} \cdot \prod q \end{gather*}\]

where the left product represents the product of all \(n\)-fold zeros of \(p\), where each \((x-a_i)\) is irreducible and the right product represents the remaining factors of \(p\). Additionally,

\[\begin{gather*} \prod_i (x-a_i)^{ni} | p \end{gather*}\]

by construction. Hence, the degree of the left side has to be smaller or equal to the degree of the right side, otherwise \(p\) would be zero. This means the degree of the left side is smaller or equal to \(m\). Since the left side represents the amount of zeros in \(p\) this means the number of zeros in \(p\) has to be less or equal to \(m\).

92) Let \(R\) be a ring and \((I_j)_{j \in J}\) be a family of ideals of \(R\). Prove that \(\bigcap_{j \in J} I_j\) is an ideal of \(R\).

We recall:

\(I\) is an ideal of \(R\) iff:

1. \(I \neq \emptyset\): Since \(0\) is contained in any ideal by definition, it will also be in the cut of ideals.
2. \(\forall x,y \in I: x -y \in I\): Since \(\forall j \in J: I_j\) is an ideal: by definition of the cut operation \(\forall a,b \in I_j\), \(a +(-b) \in I_j\) as \(I_j\) is an ideal. As this holds for any \(I_j\), \(a+ (-b) \in I\).
3. \(\forall i \in I, r \in R: i \cdot r \in I \land r \cdot i \in I\) Assume an arbitrary \(r \in R\) and \(i \in I\), then forall \(j \in J, x \in I_j\) and \(r \cdot x \in I_j\) by \(I_j\) being an ideal. Thus \(r \cdot i \in I\) by definition of the cut operation. The same argument holds for \(I \cdot r\).

93) Let \(R\) be a ring and \(I,J\) two ideals of \(R\). Does this imply that \(I \cup J\) is an ideal of \(R\)?

No, we provide a counterexample:

Let \(R = (\mathbb{Z}, +)\), \(I = 2 \mathbb{2}\) and \(J = 3 \mathbb{Z}\), then \(2 \in I\) thus \(2 \in I \cup J\), and \(3 \in J\), thus \(3 \in I \cup J\).

But \(2+3 \not \in 2\mathbb{Z} \cup 3\mathbb{Z}\) since \(5 \not \in 2\mathbb{Z}\) and \(5 \not \in 3\mathbb{Z}\).

94) Let \(R\) be a ring and \(I\) be an ideal of \(R\). Then \((R/ I,+)\) is the quotient group of \((R,+)\) over \((I,+)\). Define a multiplication on \(R/I\) by

\[\begin{gather*} (a+I)\cdot (b+I) := (ab)+I \end{gather*}\]

Prove that this operation is well-defined, i.e. that

\[\begin{gather*} \left. \begin{array}{lr} &a+I = c+I \\ \text{and } &b+I = d+I \end{array} \right\} \Longrightarrow (ab) + I = (cd) +I. \end{gather*}\]

Assume, \(a,b,c,d\) such that \(a + I = c + I\) and \(b + I = d + I\).

By \(a + I = c + I\), we know \((a+I) - (c+I) = I\) (since \(I\) is the neutral element in the quotient group).

But additionally by definition of addition in \(R/I\), \((a+I) - (c+I) = (a-c) + I\) and \((a-c) + I = I \Leftrightarrow (a-c) \in I\). Thus \(a-c \in I\) and similarly \((b-d) \in I\). Now by ideal properties: \(b(a-c) \in I\) and \(c(b-d) \in I\) and by subgroup properties: \(b(a-c) + c(b-d) \in I\), i.e. \(ba-bc+bc-bd = ab - bd \in I\). Therefore, \(ab + I = cd + I\).

Furthermore, show that \((R/I,+,\cdot)\) is a ring.

• \((R/I, +)\) is abelian: This follows directly from \((R,+)\) being abelian and \(I\) being a normal subgroup.

• \((R/I, \cdot)\) is a semigroup, i.e. satisfies associativity: We show associativity by repeatedly applying the definitions of the respective \(+\) and \(\cdot\) operations:

  \[\begin{align} &((a+I) \cdot (b+I)) \cdot (c+I) \\ =& (ab + I) \cdot (c+I) \\ =& abc + I \\ =& (a+I) \cdot (bc + I) \\ =& (a+I) \cdot ((b+I) \cdot (c+I)) \\ \end{align}\]

• distributivity: We show distributivity by repeatedly applying the definitions of the respective \(+\) and \(\cdot\) operations:

  \[\begin{align} &(a+I) \cdot ((b+I) + (c +I )) \\ =& (a+I) \cdot ((c+b) + I) \\ =& a(c+b) + I \\ =& (ac +ab) + I \\ =& (ac + I) + (ab + I) \\ =& (a+I) \cdot (c + I) + (a + I) \cdot (b+I) \\ \end{align}\]

95) Let \(U = \{\bar0, \bar2, \bar4\} \subseteq \mathbb{Z}_6\). Show that \(U\) is an ideal \(\mathbb{Z}_6\).

Let \(R = (\mathbb{Z}_6,+,\cdot)\): We recall, \(U\) is an ideal of \(R\) iff

1. \(U \neq \emptyset\)

   \(U \neq \emptyset\)

2. \(\forall x,y \in U: x + (-y) \in U\)

3. \(\forall u \in U, r \in R: u \cdot r \in U \land r \cdot u \in U\)

All conditions are satisfied, thus \(U\) is an ideal of \(R\).

Is it a subring as well?

Conditions 1) and 2) show \(U\) is a subgroup of \(R\). In 3) we showed \(U\) is closed under multiplication.

Does it have a \(1\)-element?

In 3) one can see the multiplicative identity, namely \(\bar 4\).

96) Show that \((\mathbb{Z}[x], +, \cdot)\) is a ring and that \(1 \neq (\{x, x+2\})\).

Let \(i\) be an arbitrary element such that \(i \in (\{x, x+2\})\), we can rewrite \(i = i_1 x + i_2(x+2)\). Now let \(i_2 = i_2'x + c\), where \(c\) is some constant factor, this means

\[\begin{gather*} p = p_1 x + (p_2'x+c)(x+2) = p_1 x + p_2'x^2 + cx + 2p_2'x + 2c \end{gather*}\]

we can partition this product into factors of \(x\) and constant factors:

\[\begin{gather*} p = x(p_1 + p_2' + 2p_2' x + c) + 2c \end{gather*}\]

Now, assume \(p = 1\), then \((p_1 + p_2' + 2p_2' x + c) = 0\) has to hold, but \(2c \neq 1\). Contradiction!

Remark: It can be showen that a principal deal which is generated by \(a_1,a_2,\dots,a_k\) can be alternatively generated by \(gcd(a_1,a_2, \dots, a_k)\). Therefore this example shows that \(\mathbb{Z}[x]\) is a ring where not every ideal is a principal ideal. As a consequence, $[x] cannot be a Euklidian ring.

97) Prove: If \((R,+,\cdot)\) is a ring and \(I_1\), \(I_2\) two of its ideals, then

\[\begin{align} I_1 + I_2 &:= \{a+b| a \in I_1, b \in I_2\} \text{, and} \\ I_1 * I_2 &:= \{a_1 b_1 + a_2 b_2 + \cdots + a_n b_n | n \geq 1, a_i \in I_1, b_i \in I_2 \text{ for } 1 \leq i \leq n\} \end{align}\]

are ideals as well.

Let \(I' = I_1 + I_2\), then:

1. \(I' \neq \emptyset\) by \(I_1 \neq \emptyset\) and \(I_2 \neq \emptyset\), \(I' \neq \emptyset\) by construction.
2. \(\forall x,y \in I': x + (-y) \in I'\) we can write \(x\) as \(a_1+b_1\) and \(y\) as \(a_2 + b_2\), then \(x + (-y) = (a_1+b_1) - (a_2 + b_2) = a_1 -a_2 + b_1 - b_2\), and since \(a_1,a_2 \in I_1\) , \(a_1 - a_2 \in I_1\) by \(I_1\) being an ideal. Similarly, since \(b_1,b_2 \in I_2\) , \(b_1 - b_2 \in I_2\) by \(I_2\) being an ideal. Therefore, by definition \((a_1 - a_2) + (b_1 - b_2) \in I'\).
3. \(\forall i \in I', r \in R: i \cdot r \in I' \land r \cdot i \in I'\) Assume arbitrary \(r \in R, i \in I'\), we can rewrite \(i\) by \((a+b)\) for some \(a \in I_1, b \in I_2\). Then \(r \cdot i = r \cdot (a+b) = r \cdot a + r \cdot b\) and since \(a \in I_1, r \in R\), \(r \cdot a \in I_1\) by \(I_1\) being an ideal, similarly, since \(b \in I_2, r \in R\), \(r \cdot b \in I_2\) by \(I_2\) being an ideal. Therefore, \(r \cdot a + r \cdot b \in I'\) by definition.

Since all necessary conditions hold, we conclude \(I'\) is an ideal.

Let \(I'' = I_1 * I_2\), then:

1. \(I' \neq \emptyset\) by \(I_1 \neq \emptyset\) and \(I_2 \neq \emptyset\), \(I' \neq \emptyset\) by construction.
2. \(\forall x,y \in I'': x + (-y) \in I''\) we can write \(x\) as \(a_1 b_1+ \cdots + a_n b_n\) and \(y\) as \(a'_1b'_1 + \cdots + a'_m b'_m\). W.l.o.g. assume \(n > m\): then \(x-y = (a_1 b_1 - a'_1b'_1 + \cdots + a'_m b'_m - a_m b_m - \dots - a_n b_n)\) then each summand takes one of two forms:
   • \(a_ib_i\), where \(a_i \in I_1\) and \(b_i \in I_2\), or
   • \((-a_i)b_i\), where since \(a_i \in I_1\) and \(0 \in I_1\) (by definition), \(-a_i \in I_1\) and \(b_i \in I_2\). Thus, \((a_1 b_1 - a'_1b'_1 + \cdots + a'_m b'_m - a_m b_m - \dots - a_n b_n)\) is in \(I''\).
3. \(\forall i \in I'', r \in R: i \cdot r \in I'' \land r \cdot i \in I''\) Assume arbitrary \(r \in R, i \in I'\), we can rewrite \(i\) by \(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n\) and \(r \cdot i\) by \(r(a_1 b_1) + \cdots + r(a_n b_n)\). Now, for any \(a_ib_i\), \(1 \leq i \leq n\): \(r \cdot a_i\) has to hold for any \(r \in R\) by \(I_1\) being an ideal, therefore \(r a_1 b_1 + \dots + ra_n b_n \in I''\). Since all necessary conditions hold, we conclude \(I''\) is an ideal.

98) Show that the set \(R = \{a+b\sqrt{2} | a,b \in \mathbb{Z}\}\) with the usual addition and multiplication is an integral domain but not a field.

An integral domain is a commutative ring with \(1\)-element and no zero divisor

• Commutative ring: We show \(R \leq \mathbb{R}\): * \(R \neq \emptyset\) trivially by \(\mathbb{Z} \neq \emptyset\) * \(\forall x,y \in R: x-y \in R\): we can write \(x = a + b \sqrt{2}\) and \(y = a' + b' \sqrt{2}\) then: \(x-y = a + b \sqrt{2} - (a' + b' \sqrt{2}) = a + b\sqrt{2} - a' - b'\sqrt{2} = a-a' + (b-b')\sqrt{2}\), where \(a-a' \in \mathbb{Z}\) and \(b-b'\in \mathbb{Z}\). * \(\forall x,y \in R: x \cdot y \in R\): we again can write \(x = a + b \sqrt{2}\) and \(y = a' + b' \sqrt{2}\) then: \(x \cdot y = (a + b \sqrt{2}) \cdot (a' + b'\sqrt{2}) = aa' + ab'\sqrt{2} + a'b\sqrt{2} + 2bb' = (aa'+2bb') + (ab' + a'b)\sqrt{2}\), where \((aa'+2bb') \in \mathbb{Z}\) and \((ab' + a'b) \in \mathbb{Z}\). Thus, \(R\) is a ring, and \(R\) is commutative by commutativity of \(\mathbb{R}\).

• \(1\)-element: The one element is inherited from \(\mathbb{R}\), namely \(1 + 0 \sqrt{2} = 1\).

• No zero divisor: Assume there exists a non-zero zero divisor \(a \in R\), meaning there is some \(x \in R\) such that \(ax = 0\) and \(x \neq 0\). We define a homomorphism \(\varphi: R \mapsto \mathbb{Z}\), where \(\varphi(a+b\sqrt{2}) = a^2 -2b^2\). We show \(\varphi(x) \cdot \varphi(y) = \varphi(xy)\): Let \(x = a'^2 - 2b'^2\) and \(y = a''^2 - 2b''^2\), then: \(\varphi(x) \cdot \varphi(y) = a'^2a''^2 - 2a'^2b''^2 - 2a''^2b'^2 + 4b'^2b''^2\) and: \(\varphi(xy) = (a'a'' + 2b'b'')^2 - 2(a'b'' + a''b')^2 =\) \(a'^2a''^2 + 4a'a''b'b'' + 4 b'^2b''^2 - 2a'^2b''^2 - 4a'b''a''b' - 2a''^2b'^2 =\) \(a'^2a''^2 - 2a'^2b''^2 - 2a''^2b'^2 + 4b'^2b''^2 = \varphi(x) \cdot \varphi(y)\)

  Then \(ax = 0\) has to hold, and \(\varphi(ax) = \varphi(0)\) has to be satisfied, now by \(\varphi\) being a homomorphism: \(\varphi(a) \cdot \varphi(x) = \varphi(0) = 0\), but \(\varphi(y) = 0\) only holds for \(y = 0\), thus either \(a = 0\) or \(x = 0\). Contradiction!

• But not a field, i.e.: \(\exists a \neq 0\) such that \(a\) has no multiplicative inverse. \(\varphi(2) = 4\), \(\varphi(1) = 1\) \(\varphi(2x)\) = \(\varphi(2) \cdot \varphi(x) = 4 \varphi(x)\) Now if \(2\) was unit: \(\varphi(2x) = 1\) would have to hold, but \(\varphi(2x) = 4\varphi(x) = 1\) cannot be the case since \(\varphi(x) \in \mathbb{Z}\).

Furthermore, prove that there are infinitely many units in \(R\) and give three concrete examples.

99) Find all irreducible polynomials of \(\mathbb{R}[x]\).

Hint: Show first that, if a polynomial \(p(x) \in \mathbb{R}[x]\) has a complex zero \(a+bi\), then its conjugate \(a-bi\) is a zero of \(p(x)\), too. Use this fact to conclude that every polynomial of degree at least \(3\) is reducible in \(\mathbb{R}[x]\).

100) Show that \(\sqrt{2} + \sqrt{3}\) is algebraic over \(\mathbb{Q}\) and determine its minimal polynomial.

Note the definition of algebraic: Let \(P(x)\) be monic and irreducible over \(K\), with degree \(deg(P) = n\) and \(P(a) = 0\). Then \(a\) is algebraic over \(K\) and \(P(x)\) is a minimal polynomial of \(a\) over \(K\).

Thus, we determine whether \(a = \sqrt{2} + \sqrt{3}\) is algebraic over \(\mathbb{Q}\) by:

\[\begin{align} x &= \sqrt{2} + \sqrt{3}\\ x^2 &= (\sqrt{2} + \sqrt{3})^2 = 2 + 2 \sqrt{6} + 3\\ x^2 - 5 &= 2 \sqrt{6}\\ (x^2 - 5)^2 &= (2 \sqrt{6})^2 = 4 \cdot 6\\ x^4 -10x^2 + 25 &= 24\\ x^4 -10x^2 + 1 &= 0\\ \end{align}\]

Thus \(f(x) = x^4 -10x^2 + 1\) has \(a\) as a zero in \(\mathbb{Q}\), meaning \(a\) is algebraic over \(\mathbb{Q}\).

Additionally \(f(x)\) is the minimal polynomial of \(a\) over \(\mathbb{Q}\), since the minimal polynomial of \(a\) has to have degree at least \(4\), and it is already monic (the coefficient of the largest degree component is \(1\)).